# NCERT Solutions Class 11 Chemistry

The subject of Science gets divided into three main branches of Physics, Chemistry, and Biology after Class 6 and continues to remain one of the most important subjects. Classes 11 and 12 form two of the most crucial years in terms of marks in this subject as these years will determine college admissions and future career growth.

Though the CBSE syllabus is student-friendly, there are sections that students might find a bit more challenging than the others. This is why the Board itself has recommended the regular use of NCERT solutions Class 11 Chemistry for better understanding of each concept. These solutions are easily downloadable comprehensive guides consisting of detailed explanations for each topic in plain and simple language. Since these have been compiled by subject matter experts, you can be sure that they will sufficiently aid your exam preparations.

## Chemistry NCERT Solutions for Class 11 – Chapter Wise

### Chapter 1: Some Basic Concepts of Chemistry – Term I

This chapter explains to students about the importance of chemistry, molecular mass and atomic mass. Few basic theories and laws like Avogadro’s law, Dalton’s atomic theory and the law of conservation of mass are explained in brief under this chapter. You will also solve problems based on determining the molecular weight of compounds, mass percent and concentration. You will get a clear idea about the empirical and molecular formulae, molarity, molality and mole concept.

Topics Covered in Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry for First Term:

Some Basic Concepts of Chemistry: General Introduction: Importance and scope of Chemistry. Atomic and molecular masses, mole concept and molar mass, percentage composition, empirical and molecular formula, chemical reactions, stoichiometry and calculations based on stoichiometry.

### Chapter 2: Structure of Atom – Term I

Students will learn about Thomson’s atomic model, subatomic particles, Bohr’s model, Rutherford’s atomic model and quantum mechanical model of atom. Problems on relationship between frequency and wavelength, energy associated with electromagnetic radiation and subatomic particles are present in this chapter. You will get to know how to write the electron configurations and the electron transition in different shells which is important for the exam.

Topics Covered in Class 11 Chemistry Chapter 2 Structure of Atom for First Term:

Structure of Atom: Bohr’s model and its limitations, concept of shells and subshells, dual nature of matter and light, de Broglie’s relationship, Heisenberg uncertainty principle, concept of orbitals, quantum numbers, shapes of s, p and d orbitals, rules for filling electrons in orbitals – Aufbau principle, Pauli’s exclusion principle and Hund’s rule, electronic configuration of atoms, stability of half-filled and completely filled orbitals.

### Chapter 3: Classification of Elements and Periodicity in Properties – Term I

From the exam point of view, classification of elements is very important. Most of the questions in the final exam and various competitive exams appear from this chapter. So learning all the concepts from this chapter is important. From this chapter, students will also learn about the s-block, p-block, d-block and f-block elements from the periodic table, trends in physical and chemical properties and chemical reactivity. By learning this chapter thoroughly, students will obtain a good score in the exam.

Topics Covered in Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Properties for First Term:

Classification of Elements and Periodicity in Properties: Modern periodic law and the present form of periodic table, periodic trends in properties of elements -atomic radii, ionic radii, inert gas radii, Ionization enthalpy, electron gain enthalpy, electronegativity, valency. Nomenclature of elements with atomic number greater than 100.

### Chapter 4: Chemical Bonding and Molecular Structure – Term I

Chemistry is based on the things which we observe in our surroundings and is called the central science. The topics discussed in this chapter are – VSEPR Theory, Lewis structures, Valence Bond Theory, polar character of covalent bonds and hydrogen bonding. Students will learn how to draw Lewis dot symbols for molecules, atoms and polyatomic ions. The solutions at BYJU’S contain diagrams for each concept to provide visual learning experience for the students.

Topics Covered in Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure for First Term:

Chemical Bonding and Molecular Structure:

Valence electrons, ionic bond, covalent bond, bond parameters, Lewis structure, polar character of covalent bond, covalent character of ionic bond, valence bond theory, resonance, geometry of covalent molecules, VSEPR theory, concept of hybridization, involving s, p and d orbitals and shapes of some simple molecules, molecular orbital theory of homonuclear diatomic molecules(qualitative idea only), Hydrogen bond.

### Chapter 5: States of Matter – Term II

This chapter explains the intermolecular forces and how they affect the physical state of a substance. It deals with other concepts associated with gaseous and liquid states of matter. This is the main reason why this chapter is important for the exam. Students will be able to understand the Boyle’s Law, Charle’s Law, Gay Lussac’s Law and Avogadro Law which carries more marks as per the exam pattern. Problems on finding partial pressure, critical temperature and pressure, Van der Waals force and other intermolecular forces are present in this chapter.

Topics Covered in Class 11 Chemistry Chapter 5 States of Matter for Second Term:

States of Matter: Gases and Liquids: Three states of matter, intermolecular interactions, types of bonding, melting and boiling points, role of gas laws in elucidating the concept of the molecule, Boyle’s law, Charles law, Gay Lussac’s law, Avogadro’s law, ideal behaviour, empirical derivation of gas equation, Avogadro’s number, ideal gas equation and deviation from ideal behaviour.

### Chapter 6: Thermodynamics – Term II

It is a branch of science which explains the relationship between heat and the other energy forms. On the basis of exchange of matter and energy, thermodynamic systems are divided into three types – open, closed and isolated system. Students will learn more about the concepts like – terms in thermodynamics, applications, calorimetry, enthalpies, spontaneity and Gibbs energy change and equilibrium. The problems in the NCERT Solutions are answered in a systematic manner completely based on the latest syllabus of the CBSE board.

Topics Covered in Class 11 Chemistry Chapter 6 Thermodynamics for Second Term:

Chemical Thermodynamics: Concepts of System and types of systems, surroundings, work, heat, energy, extensive and intensive properties, state functions.

First law of thermodynamics -internal energy and enthalpy, measurement of U and H, Hess’s law of constant heat summation, enthalpy of bond dissociation, combustion, formation, atomization, sublimation, phase transition, ionization, solution and dilution. Second law of Thermodynamics (brief introduction) Introduction of entropy as a state function, Gibb’s energy change for spontaneous and nonspontaneous processes.

### Chapter 7: Equilibrium – Term II

From this chapter, students will learn buffer solutions, equilibrium constant and the common ion effect. As this chapter contains a lot of concepts, thorough understanding of each of them is important to perform well in the annual exam. The solutions contain clear cut explanations for each and every concept so that all the queries of the students are clarified. The concepts discussed in this chapter are – solid liquid equilibrium, applications of equilibrium constant, factors affecting equilibria, acids, bases and salts, buffer solutions etc.

Topics Covered in Class 11 Chemistry Chapter 7 Equilibrium for Second Term:

Equilibrium: Equilibrium in physical and chemical processes, dynamic nature of equilibrium, law of mass action, equilibrium constant, factors affecting equilibrium – Le Chatelier’s principle, ionic equilibrium- ionization of acids and bases, strong and weak electrolytes, degree of ionization, ionization of poly basic acids, acid strength, concept of pH, buffer solution, solubility product, common ion effect (with illustrative examples).

### Chapter 8: Redox Reactions – Term I

The fact here is that the field of electrochemistry deals with redox reactions. It is important for the students to learn this chapter in order to score good marks in the final exam. Students will get questions based on the classical idea of redox reaction, redox reactions in terms of electron transfer reactions, oxidation number and electrode processes. Students are recommended to first complete the entire chapter using the NCERT textbook prescribed by the CBSE board.

Topics Covered in Class 11 Chemistry Chapter 8 Redox Reaction for First Term:

Redox Reactions:

Concept of oxidation and reduction, redox reactions, oxidation number, balancing redox reactions, in terms of loss and gain of electrons and change in oxidation number.

### Chapter 9: Hydrogen – Term I

Hydrogen is the lightest element with one electron and proton. The abundant element in the entire universe is dihydrogen. Hydrogen is an important element which makes up 80% of the entire mass of the universe. It is very important for the students to know about this element in order to secure good marks in the final exam. Students will also learn about preparation and properties of dihydrogen and its applications in our daily life. It might be difficult for the students to grasp all these concepts at a stretch, so making use of a perfect study material will help you through it.

Topics Covered in Class 11 Chemistry Chapter 9 Hydrogen for First Term:

Hydrogen: Position of hydrogen in periodic table, occurrence, isotopes, hydrides-ionic covalent and

interstitial; physical and chemical properties of water, heavy water, hydrogen as a fuel.

### Chapter 10: The s-Block Elements – Term II

Students will learn about the Alkali metals and Alkaline earth metals which are the group 1 and group 2 in the periodic table. Some of the concepts like physical and chemical properties of s-block elements, general characteristics of compounds and about important compounds like calcium. The biological importance of calcium and magnesium is also explained clearly in this chapter. Periodic table and the elements present in them is a hot topic from the exam perspective. For this reason, students are highly recommended to learn all these concepts thoroughly to score well.

Topics Covered in Class 11 Chemistry Chapter 10 The s-Block Elements for Second Term:

s -Block Elements: Group 1 and Group 2 Elements -General introduction, electronic configuration,

occurrence, anomalous properties of the first element of each group, diagonal relationship, trends in

the variation of properties (such as ionization enthalpy, atomic and ionic radii), trends in chemical

reactivity with oxygen, water, hydrogen and halogens, uses.

### Chapter 11: The p-Block Elements – Term II

The elements present between the 13th and 18th group are called the p-block elements. As this chapter has a lot of concepts, it is very important for the students to revise it regularly. Students will get a gist of the p-block elements, where the last electron enters or is found on the p-subshell. Some of the concepts which you will learn in this chapter are – boron family, important compounds of boron, uses of aluminium and boron, carbon family, allotropes of carbon and important compounds of silicon and carbon.

Topics Covered in Class 11 Chemistry Chapter 11 The p-Block Elements for Second Term:

Some p -Block Elements: General Introduction to p -Block Elements

Group 13 Elements: General introduction, electronic configuration, occurrence, variation of properties, oxidation states, trends in chemical reactivity, anomalous properties of first element of the group, Boron – physical and chemical properties.

Group 14 Elements: General introduction, electronic configuration, occurrence, variation of properties, oxidation states, trends in chemical reactivity, anomalous behaviour of first elements.

Carbon-catenation, allotropic forms, physical and chemical properties.

### Chapter 12: Organic Chemistry – Some Basic Principles and Techniques – Term I

When it comes to exam preparation, both organic and inorganic chemistry plays an important role in attaining good marks. This chapter will introduce you to concepts such as organic compounds, structural representation, classification, nomenclature, isomerism, purification of organic compounds and quantitative analysis. If students are not able to learn these concepts using the NCERT textbook they can download the solutions which are available online for absolutely free of cost.

Topics Covered in Class 11 Chemistry Chapter 12 Organic Chemistry – Some Basic Principles for First Term:

Organic Chemistry: Some basic Principles and Techniques: General introduction, classification and IUPAC nomenclature of organic compounds. Electronic displacements in a covalent bond: inductive effect, electrometric effect, resonance and hyperconjugation. Homolytic and heterolytic fission of a covalent bond: free radicals, carbocations, carbanions, electrophiles and nucleophiles, types of organic reactions.

### Chapter 13: Hydrocarbons – Term II

The scientific study of properties, structure, reactions, composition and synthesis of organic compounds is called organic chemistry. It teaches the students about concepts like alkanes and its preparation from alkyl halides, unsaturated hydrocarbons and carboxylic acids. Students will also learn about the physical and chemical properties of alkanes as per the prescribed CBSE syllabus. As this chapter is important from the exam point of view, students must highly concentrate on all the concepts and memorise it thoroughly to perform well in the exam.

Topics Covered in Class 11 Chemistry Chapter 13 Hydrocarbons for Second Term:

Hydrocarbons: Classification of Hydrocarbons Aliphatic Hydrocarbons:

Alkanes – Nomenclature, isomerism, conformation (ethane only), physical properties, chemical reactions.

Alkenes – Nomenclature, structure of double bond (ethene), geometrical isomerism, physical properties, methods of preparation, chemical reactions: addition of hydrogen, halogen, water, hydrogen halides (Markovnikov’s addition and peroxide effect), ozonolysis, oxidation, mechanism of electrophilic addition.

Alkynes – Nomenclature, structure of triple bond (ethyne), physical properties, methods of preparation, chemical reactions: acidic character of alkynes, addition reaction of – hydrogen, halogens, hydrogen halides and water.

Aromatic Hydrocarbons: Introduction, IUPAC nomenclature, benzene: resonance, aromaticity, chemical properties: mechanism of electrophilic substitution. Nitration, sulphonation, halogenation, Friedel Craft’s alkylation and acylation, directive influence of functional group in monosubstituted benzene. Carcinogenicity and toxicity.

### Chapter 14: Environmental Chemistry

Various concepts which are related to the environment like water pollution, atmospheric pollution, soil pollution and its reasons are explained briefly in this chapter. Along with this students will also learn about the use of pesticides to reduce soil pollution and strategies to control pollution like waste disposal. By learning this chapter thoroughly, students will get to know about the pollution occurring in nature by our activities.

NCERT Chemistry Class 11

• Brief explanations are provided for each question.
• The numerical equations are solved in a stepwise manner.
• The answers are clear cut and precise completely based on the CBSE syllabus.
• Subject matter experts design the solutions after conducting a vast research.

Q.1 Predict which of the following reaction will have appreciable concentration of reactants and products:

$\begin{array}{l}a\right)\text{}C{l}_{2}\left(g\right)\underset{}{⇌}2Cl\left(g\right),\text{}{K}_{c}=5×{10}^{-39}\\ b\right)\text{}C{l}_{2}\left(g\right)+2NO\left(g\right)\underset{}{⇌}NOCl\left(g\right),\text{}{K}_{c}=3.7×{10}^{8}\\ c\right)\text{}C{l}_{2}\left(g\right)+2N{O}_{2}\left(g\right)\underset{}{⇌}2N{O}_{2}Cl\left(g\right),\text{}{K}_{c}=1.8\end{array}$

Ans.

For the reaction (c), Kc is neither high nor very low. Therefore, in this reaction, reactants and products will be present in appreciable concentration.

Q.2 Calculate the atomic mass (average) of chlorine using the following data:

 % Natural Abundance Molar Mass 35Cl 75.77 34.9689 37Cl 24.23 36.9659

Ans.

The average atomic mass of chlorine
(Fractional abundance of 35Cl)(Molar mass of 35Cl)+(Fractional abundance of 37Cl)
(Molar mass of 37Cl)=[{(75.77100)(34.9689)}+{(24.23100) (36.9659 u)}]
= 26.4959 + 8.9568
= 35.4527 u
The average atomic mass of chlorine = 35.4527 u

Q.3 In three moles of ethane (C2H6), calculate the following:

(i) Number of moles of carbon atoms.

(ii) Number of moles of hydrogen atoms.

(iii) Number of molecules of ethane.

Ans.

(i) 1 mole of C2H6 contains 2 moles of carbon atoms.

Number of moles of carbon atoms in 3 moles of C2H6

= 2 × 3 = 6

(ii) 1 mole of C2H6 contains 6 moles of hydrogen atoms.

Number of moles of carbon atoms in 3 moles of C2H6

= 3 × 6 = 18

(iii) 1 mole of C2H6 contains 6.023 × 1023 molecules of ethane.

Number of molecules in 3 moles of C2H6

= 3 × 6.023 × 1023 = 18.069 × 1023

Q.4 The value of Kc for the reaction

$3{O}_{2}\left(g\right)\underset{}{⇌}2{O}_{3}\left(g\right)$

is 2.0 × 10–50 at 25°C. If the equilibrium concentration of O2 in air at 25°C is 1.6 × 10–2, what is the concentration of O3?

Ans.

The given reaction is:

$\begin{array}{l}3{O}_{2}\left(g\right)\underset{}{⇌}2{O}_{3}\left(g\right)\\ {K}_{c}=\frac{{\left[{O}_{3}\left(g\right)\right]}^{2}}{{\left[{O}_{2}\left(g\right)\right]}^{3}}\end{array}$

It is given that Kc= 2.0 ×10–50 and [O2] = 1.6 ×10–2

$\begin{array}{l}⇒2.0×{10}^{-50}=\frac{{\left[{O}_{3}\left(g\right)\right]}^{2}}{{\left(1.6×{10}^{-2}\right)}^{3}}\\ ⇒{\left[{O}_{3}\left(g\right)\right]}^{2}=\left(2.0×{10}^{-50}\right){\left(1.6×{10}^{-2}\right)}^{3}\\ ⇒{\left[{O}_{3}\left(g\right)\right]}^{2}=8.192×{10}^{-56}\\ ⇒\left[{O}_{3}\left(g\right)\right]=2.86×{10}^{28}\text{M}\end{array}$

Hence, the concentration O3 is 2.86 x 10-28 M.

Q.5 What is the concentration of sugar (C12H22O11) in mol L-1 if its 20 g are dissolved in enough water to make a final volume up to 2 L?

Ans.

Molarity (M) of a solution is given by,

$\begin{array}{l}=\frac{\text{Number of moles of solute}}{\text{volume of solution in Litres}}\\ =\frac{\text{Mass ofsugar / molar massofsugar}}{\text{2 L}}\\ =\frac{20g/\left[\left(12×12\right)+\left(1×22\right)+\left(11×16\right)\right]g}{2L}\\ =\frac{20g/342g}{2L}\\ =\frac{0.0585mol}{2L}\end{array}$

= 0.02925 mol L–1
Molar concentration of sugar = 0.02925 mol L–1

Q.6 If the density of methanol is 0.793 kg L–1, what is its volume needed for making 2.5 L of its 0.25 M solution?

Ans.

Molar mass of methanol (CH3OH) = (1 × 12) + (4 × 1) + (1 × 16)

= 32 g mol–1

= 0.032 kg mol–1

$\text{Molarity of given methanol}=\frac{0.793kg{L}^{-1}}{0.032kgmo{l}^{-1}}$

Molarity of methanol solution = 24.78 mol L–1
(Since density is mass per unit volume)
Applying,
M1V1 = M2V2
(Given solution) (Solution to be prepared)
(24.78 mol L-1) V1 = (2.5 L) (0.25 mol L-1)
V1 = 0.0252 L
V1 = 25.22 mL

Q.7 The reaction,

$CO\left(g\right)+3{H}_{2}\left(g\right)\underset{}{⇌}C{H}_{4}\left(g\right)+{H}_{2}O\left(g\right)$

is at equilibrium at 1300 K in a 1L flask. It also contain 0.30 mol of CO, 0.10 mol of H2 and 0.02 mol of H2O and an unknown amount of CH4 in the flask. Determine the concentration of CH4 in the mixture. The equilibrium constant, Kc for the reaction at the given temperature is 3.90.

Ans.

Equilibrium

Conc. 0.30 0.10 x 0.02

$CO\left(g\right)+3{H}_{2}\left(g\right)\underset{}{⇌}C{H}_{4}\left(g\right)+{H}_{2}O\left(g\right)$

The equilibrium constant for the reaction,

$\begin{array}{l}{K}_{c}=\frac{\left[C{H}_{4}\left(g\right)\right]\left[{H}_{2}O\left(g\right)\right]}{\left[CH\left(g\right)\right]{\left[{H}_{2}O\left(g\right)\right]}^{3}}\\ ⇒\frac{x×0.02}{0.3×{\left(0.1\right)}^{3}}=3.90\\ ⇒x=\frac{3.90×0.3×{\left(0.1\right)}^{3}}{0.02}\end{array}$

or, x = 0.0585 M = 5.85 x 10-2 M

Hence, the concentration of CH4 at equilibrium is 5.85 × 10–2 M.

Q.8 What is meant by the conjugate acid/base pair? Find the conjugate acid/base for the following species:

HNO2, CN, HClO4, F, OH, CO32- and S2-

Ans.

The acid-base pair which differs by a proton is called conjugate acid/base pair.

 Species Conjugate acid/base HNO2 NO2– (base) CN– HCN (acid) HClO4– ClO4– (base) F– HF (acid) OH– H2O(acid)/ O2– (base) CO32– HCO3– (acid) S2– HS– (acid)

Q.9 Pressure is determined as force per unit area of the surface. The SI unit of pressure, Pascal is as shown below:

1Pa = 1N m–2

If mass of air at sea level is 1034 g cm–2, calculate the pressure in Pascal.

Ans.

Pressure is defined as force acting per unit area of the surface.

$P=\frac{F}{A}$ $=\frac{1034g×9.8m{s}^{-2}}{c{m}^{2}}×\frac{1kg}{1000g}×\frac{{\left(100\right)}^{2}c{m}^{2}}{1{m}^{2}}$

= 1.01332 × 105 kg m–1s–2

We know,

1 N = 1 kg ms–2

Then,

1 Pa = 1 Nm–2 = 1 kg m–2s–2

1 Pa = 1 kg m–2s–2

Pressure = 1.01332 × 105Pa

Q.10 Which of the followings are Lewis acids? H2O, BF3, H+ and NH4+.

Ans.

Lewis acid can accept a pair of electrons. All cations are Lewis acid. Thus, BF3, H+ and NH4+ are considered as Lewis acids.

Q.11 What is the SI unit of mass? How is it defined?

Ans.

The SI unit of mass is kilogram (kg). 1 Kilogram is defined as the mass equal to the mass of the international prototype of kilogram.

Q.12 What will be the conjugate bases for the Bronsted acids: HF, H2SO4 and HCO3?

Ans.

The table below lists the conjugate bases for the given Bronsted acids.

 Bronsted acids Conjugate bases HF F– H2SO4 HSO4– HCO3– CO32–

Q.13 Write the conjugate acids for the following Bronsted bases: NH2, NH3 and HCOO.

Ans.

The table below lists the conjugate acids for the given Bronsted bases:

 Bronsted bases Conjugate acids NH2– NH3 NH3 NH4+ HCOO– HCOOH

Q.14 The species: H2O, HCO3, HSO4 and NH3 can act both as Bronsted acids and bases. For each case give the corresponding conjugate acid and base.

Ans.

The table below lists the conjugate acids and conjugate bases for the given species.

 Species Conjugate acids Conjugate bases H2O H3O+ OH– HCO3– H2CO3 CO32– HSO4– H2SO4 SO42– NH3 NH4+ NH2–

Q.15 Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/base:

1. OH (b) F (c) H+ and (d) BCl3

Ans.

1. OH acts as Lewis base as it can donate its lone pair of electrons.
2. Fis Lewis base as it can donate one of its lone pair of electrons.
3. H+ acts as Lewis acid since it can accept a lone pair of electrons.
4. BCl3 is Lewis acid since it can accept a lone pair of electrons in vacant p-orbital of Boron.

Q.16 Match the following prefixes with their multiples:

 Prefixes Multiples (i) Micro 106 (ii) Deca 109 (iii) Mega 10-6 (iv) Giga 10-15 (v) Femto 10

Ans.

 Prefixes Multiples (i) Micro 10-6 (ii) Deca 10 (iii) Mega 106 (iv) Giga 109 (v) Femto 10-15

Q.17 What do you mean by significant figures?

Ans.

Significant figures are those meaningful digits that are known with certainty .

They indicate uncertainty in an experiment or calculated value. For example, if 15.6 mL is the result of an experiment, then 15 is certain while 6 is uncertain, and the total number of significant figures are 3.

Hence, significant figures are defined as the total number of digits in a number including the last digit that represents the uncertainty of the result.

Q.18 The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10–3 M. What is its pH?

Ans.

$\begin{array}{l}\left[{H}^{+}\right]={3.8510}^{-3}M\\ pH=-\mathrm{log}\left[{H}^{+}\right]\\ ⇒pH=-\mathrm{log}\left(3.8×{10}^{-3}M\right)\\ ⇒pH=-\mathrm{log}3.8-\mathrm{log}{10}^{-3}\\ ⇒pH=-\mathrm{log}3.8+3\\ =-0.58+3\\ =2.42\end{array}$

Q.19 The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it.

Ans.

Given pH= 3.76

From the relation,

$\begin{array}{l}pH=-\mathrm{log}\left[{H}^{+}\right]\\ ⇒\mathrm{log}\left[{H}^{+}\right]=-pH\\ ⇒\left[{H}^{+}\right]=anti\mathrm{log}\left(-pH\right)\\ ⇒\left[{H}^{+}\right]=anti\mathrm{log}\left(-3.76\right)\\ =1.74×{10}^{-4}M\end{array}$

Hence, the concentration of hydrogen ion in vinegar is 1.74 × 10–4 M.

Q.20 The ionization constant of HF, HCOOH and HCN at 298 K are 6.8 × 10–4, 1.8 × 10–4 and 4.8 × 10–9 respectively. Calculate the ionization constants of the corresponding conjugate base.

Ans.

From the relation,

${K}_{w}={K}_{a}×{K}_{b}$

Here,

Kw= ionic product of water

Kb = ionization constant of conjugate base

Ka = ionization constant of conjugate acid

${K}_{b}=\frac{{K}_{w}}{{K}_{a}}$

Ka of HF = 6.8 × 10–4

Kw of water = 1 x 10-14

Hence, Kb of its conjugate base F

$\begin{array}{l}=\frac{{K}_{w}}{{K}_{a}}\\ =\frac{1×{10}^{-14}}{6.8×{10}^{-4}}=1.5×{10}^{-11}\end{array}$

Again, Ka of HCOOH = 1.8 × 10–4

So, Kb of its conjugate base HCOO

$=\frac{{K}_{w}}{{K}_{a}}=\frac{1×{10}^{-14}}{1.8×{10}^{-4}}=5.6×{10}^{-11}$

Given, Ka of HCN = 4.8 × 10–9

$=\frac{{K}_{w}}{{K}_{a}}=\frac{1×{10}^{-14}}{4.8×{10}^{-9}}=2.08×{10}^{-6}$

Q.21 A sample of drinking water was found to be severely contaminated with chloroform, CHCl3, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass).

(i) Express this in percent by mass.

(ii) Determine the molality of chloroform in the water sample.

Ans.

(i) 1 ppm is equivalent to 1 part out of 1 million (106) parts.

Mass percent of 15 ppm chloroform in water

$=\frac{15}{{10}^{6}}×100$ $\simeq 1.5×{10}^{-3}%$

(ii) 100 g of the sample contains 1.5 × 10-3 g of CHCl3
⇒1000 g of the sample contains 1.5 × 10-2 g of CHCl3
∴ Molality of chloroform in water

$=\frac{1.5×{10}^{-2}g}{{\text{Molar mass of CHCI}}_{3}}$

Molar mass of CHCl3 = 12.00 + 1.00 + 3(35.5)
= 119.5 g mol-1
∴ Molality of chloroform in water = 0.0125 × 10-2 m
= 1.25 × 10-4 m

Q.22 Express the following in the scientific notation:

(i) 0.0048

(ii) 234,000

(iii) 8008

(iv) 500.0

(v) 6.0012

Ans.

(i) 0.0048 = 4.8× 10–3

(ii) 234, 000 = 2.34 ×105

(iii) 8008 = 8.008 ×103

(iv) 500.0 = 5.000 × 102

(v) 6.0012 = 6.0012

Q.23 How many significant figures are present in the following?

(i) 0.0025

(ii) 208

(iii) 5005

(iv) 126,000

(v) 500.0

(vi) 2.0034

Ans.

(i) 0.0025

There are 2 significant figures.

(ii) 208

There are 3 significant figures.

(iii) 5005

There are 4 significant figures.

(iv) 126,000

There are 3 significant figures.

(v) 500.0

There are 4 significant figures.

(vi) 2.0034

There are 5 significant figures.

Q.24 Round up the following upto three significant figures:

(i) 34.216

(ii) 10.4107

(iii) 0.04597

(iv) 2808

Ans.

(i) 34.2

(ii) 10.4

(iii) 0.0460

(iv) 2810

Q.25 The ionization constant of phenol is 1.0 × 10–10. What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization if the solution is also 0.01 M in sodium phenolate?

Ans.

Ionization of phenol involves the following reaction:

$\begin{array}{l}\underset{\begin{array}{l}\\ Initial\text{conc}\text{. 0}\text{.05}\\ \\ \text{At equilibrium}\left(0.05-x\right)\end{array}}{{C}_{6}{H}_{5}OH}+{H}_{2}O\underset{}{⇌}{\underset{\begin{array}{l}\\ \text{0}\\ \\ \text{x}\end{array}}{{C}_{6}{H}_{5}O}}^{-}+\underset{\begin{array}{l}\\ \text{0}\\ \\ \text{x}\end{array}}{{H}_{3}{O}^{+}}\\ {K}_{a}=\frac{\left[{C}_{6}{H}_{5}{O}^{-}\right]\left[{H}_{3}{O}^{+}\right]}{\left[{C}_{6}{H}_{5}OH\right]}\\ ⇒{K}_{a}\frac{x\cdot x}{\left(0.05-x\right)}\\ \text{The value of the ionization constant is very less},\text{x will be very small}.\text{We can ignore x in the denominator}.\\ x=\sqrt{\left(1×{10}^{-10}×0.05\right)}\\ x=\sqrt{\left(5×{10}^{-12}\right)}\\ =2.2×{10}^{-6}M=\left[{H}_{3}{O}^{+}\right]\\ Since\left[{H}_{3}{O}^{+}\right]=\left[{C}_{6}{H}_{5}{O}^{-}\right]\\ \left[{C}_{6}{H}_{5}{O}^{-}\right]=2.2×{10}^{-6}M\end{array}$

Let the degree of ionization of phenol be α.

$\begin{array}{l}\underset{\begin{array}{l}\\ Conc.\left(0.05-0.05\alpha \right)\text{}\end{array}}{{C}_{6}{H}_{5}OH}+{H}_{2}O\underset{}{⇌}\underset{\begin{array}{l}\\ 0.05\alpha \end{array}}{{C}_{6}{H}_{5}{O}^{-}}+\underset{\begin{array}{l}\\ 0.05\alpha \end{array}}{{H}_{3}{O}^{+}}\\ \underset{\begin{array}{l}\\ Conc.\text{}\end{array}}{{C}_{6}{H}_{5}ONa}\underset{}{⇌}{C}_{6}{H}_{5}{O}^{-}+\underset{\begin{array}{l}\\ 0.01\end{array}}{N{a}^{+}}\end{array}$

[C6H5OH]= 0.05 – 0.05α; 0.05 M

[C6H5O] = 0.01 + 0.05α; 0.01 M

[H3O+] = 0.05α

$\begin{array}{l}{K}_{a}=\frac{\left[{C}_{6}{H}_{5}{O}^{-}\right]\left[{H}_{3}{O}^{+}\right]}{\left[{C}_{6}{H}_{5}OH\right]}\\ {K}_{a}=\frac{\left(0.01\right)\left(0.05\alpha \right)}{\left(0.05\right)}\\ 1×{10}^{-10}=0.01\alpha \\ \alpha =1×{10}^{-8}\end{array}$

Q.26 The following data are obtained when dinitrogen and dioxygen react together to form different compounds:

Mass of dinitrogen Mass of dioxygen

(i) 14 g 16 g

(ii) 14 g 32 g

(iii) 28 g 32 g

(iv) 28 g 80 g

(a) Which law of chemical combination is obeyed by the above experimental data? Give its statement.

(b) Fill in the blanks in the following conversions:

(i) 1 km = …………………. mm = …………………. pm

(ii) 1 mg = …………………. kg = …………………. ng

(iii) 1 mL = …………………. L = …………………. dm3

Ans.

If we fix the mass of dinitrogen at 28 g, then the masses of dioxygen that will combine with the fixed mass of dinitrogen are 32 g, 64 g, 32 g, and 80 g. The masses of dioxygen bear a whole number ratio of 2:4:2:5. Hence, the given experimental data obeys the law of multiple proportions. The law states that if two elements combine to form more than one compound, then the masses of one element that combines with the fixed mass of another element are in the ratio of small whole numbers.

$\begin{array}{l}\left(b\right)1km=1km×\frac{1000m}{1km}×\frac{100cm}{1m}×\frac{10mm}{1cm}\\ \therefore 1km={10}^{6}mm\\ 1km=1km×\frac{1000m}{1km}×\frac{1pm}{{1}^{-12}m}\\ \therefore 1km={10}^{15}pm\\ \text{Hence},\text{1 km}=\text{1}{0}^{\text{6}}\text{mm}=\text{1}{0}^{\text{15}}\text{pm}\\ \left(ii\right)1mg=1mg×\frac{1g}{1000mg}×\frac{1kg}{1000g}\\ ⇒1mg={10}^{-6}kg\\ 1mg=1mg×\frac{1g}{1000mg}×\frac{1kg}{{10}^{-9}g}\\ ⇒1mg={10}^{6}ng\\ \therefore 1mg={10}^{-6}kg={10}^{6}ng\\ \left(iii\right)1mL=1mL×\frac{1L}{1000mL}\\ ⇒1mL={10}^{-3}L\\ 1mL=1c{m}^{3}=1c{m}^{3}×\frac{1dm}{10cm}×\frac{1dm}{10cm}×\frac{1dm}{10cm}\\ ⇒1mL={10}^{-3}d{m}^{3}\\ \therefore 1mL={10}^{-3}L={10}^{-3}d{m}^{3}\end{array}$

Q.27 If the speed of light is 3.0 ×108 m s–1, calculate the distance covered by light in 2.00 ns.

Ans.

According to the question:

Time taken to cover the distance = 2.00 ns

= 2.00 × 10–9 s

Speed of light = 3.0 × 108 ms–1

Distance travelled by light in 2.00 ns

= Speed of light × Time taken

= (3.0 × 108 ms –1) (2.00 × 10–9 s)

= 6.00 × 10–1m

= 0.600 m

Q.28 In a reaction

A + B2 → AB2

Identify the limiting reagent, if any, in the following reaction mixtures.

(i) 300 atoms of A + 200 molecules of B

(ii) 2 mol A + 3 mol B

(iii) 100 atoms of A + 100 molecules of B

(iv) 5 mol A + 2.5 mol B

(v) 2.5 mol A + 5 mol B

Ans.

A limiting reagent determines the extent of a reaction. It is the reactant which is the first to get consumed during a reaction, thereby causing the reaction to stop and limiting the amount of products formed.

(i) According to the given reaction, 1 atom of A reacts with 1 molecule of B. Thus, 200 molecules of B will react with 200 atoms of A, thereby leaving 100 atoms of A unused. Hence, B is the limiting reagent.

(ii) According to the reaction, 1 mol of A reacts with 1 mol of B. Thus, 2 mol of A will react with only 2 mol of B. As a result, 1 mol of A will not be consumed. Hence, A is the limiting reagent.

(iii) According to the given reaction, 1 atom of A combines with 1 molecule of B. Thus, all 100 atoms of A will combine with all 100 molecules of B. Hence, the mixture is stoichiometric where no limiting reagent is present.

(iv) 1 mol of atom A combines with 1 mol of molecule B. Thus, 2.5 mol of B will combine with only 2.5 mol of A. As a result, 2.5 mol of A will be left as such. Hence, B is the limiting reagent.

(v) According to the reaction, 1 mol of atom A combines with 1 mol of molecule B. Thus, 2.5 mol of A will combine with only 2.5 mol of B and the remaining 2.5 mol of B will be left as such. Hence, A is the limiting reagent.

Q.29 Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:

N2(g)+ 3H2(g) → 2NH3(g)

(i) Calculate the mass of ammonia produced if 2.00 × 103g dinitrogen reacts with 1.00 × 103 g of dihydrogen.

(ii) Will any of the two reactants remain unreacted?

(iii) If yes, which one and what would be its mass?

Ans.

(i) Balancing the given chemical equation,
N2(g)+ 3H2(g)→2NH3(g)
From the equation, 1 mole (28 g) of dinitrogen reacts with 3 mole (6 g) of dihydrogen to give 2 mole (34 g) of ammonia.
⇒ 2.00 × 103 g of dinitrogen will react with

$\frac{6g}{28g}×2.00×{10}^{3}g$

dihydrogen i.e., 2.00 × 103 g of dinitrogen will react with 428.6 g of dihydrogen.
Given,
Amount of dihydrogen = 1.00 × 103 g
Hence, N2 is the limiting reagent.
∴ 28g of N2 produces 34 g of NH3
Hence, mass of ammonia produced by 200 g of N2

$=\frac{34g}{28g}×2000g$

= 2428.57 g
(ii) N2 is the limiting reagent and H2 is the excess reagent. Hence, H2 will remain unreacted.
(iii) Mass of dihydrogen left unreacted = 1.00 × 103 g – 428.6 g
= 571.4 g

Q.30 How are 0.50 mol Na2CO3 and 0.50 M Na2CO3 different?

Ans.

Molar mass of Na2CO3 = (2 × 23) + 12.00 + (3 × 16)

= 106 g mol–1

Now, 1 mole of Na2CO3 means 106 g of Na2CO3

$\begin{array}{l}\therefore 0.5molofN{a}_{2}\mathrm{CO}{}_{3}=\frac{106g}{1mole}×0.5molofN{a}_{2}\mathrm{CO}{}_{3}\\ =53gN{a}_{2}\mathrm{CO}{}_{3}\\ ⇒0.50MofN{a}_{2}\mathrm{CO}{}_{3}=0.50mol/LofN{a}_{2}\mathrm{CO}{}_{3}\end{array}$

Hence, 0.50 mol of Na2CO3 is present in 1 L of water or 53 g of Na2CO3 is present in 1 L of water.

Q.31 If ten volumes of dihydrogen gas react with five volumes of dioxygen gas, how many volumes of water vapour would be produced?

Ans.

Reaction of dihydrogen with dioxygen can be written as:

2H2(g)+ O2(g) → 2H2O(g)

Now, two volumes of dihydrogen react with one volume of dioxygen to produce two volumes of water vapour.

Hence, ten volumes of dihydrogen will react with five volumes of dioxygen to produce ten volumes of water vapour.

Q.32 The first ionization constant of H2S is 9.1 × 10–8.

(i) Calculate the concentration of HSion in its 0.1 M solution.

(ii)How will this concentration be affected if the solution is 0.1 M in HCl?

(iii)Also if the second dissociation constant of H2S is 1.2 × 10–13, calculate the concentration of S2– under both conditions.

Ans.

(i) Let us calculate the concentration of HSion:

$\begin{array}{l}{K}_{a}=\frac{\left[{H}^{+}\right]\left[H{S}^{-}\right]}{\left[{H}_{2}S\right]}\\ ⇒9.1×{10}^{-8}=\frac{x\cdot x}{\left(0.1-x\right)}\\ ⇒\left(9.1×{10}^{-8}\right)\left(0.1-x\right)={x}^{2}\end{array}$

Since value of x is very small, so we can consider

0.1 – x = 0.1

$\begin{array}{l}⇒9.1×{10}^{-8}=\frac{{x}^{2}}{0.1}\\ ⇒{x}^{2}=9.1×{10}^{-9}\\ \left[H{S}^{-}\right]=x=9.54×{10}^{-5}M\end{array}$

(ii) In the presence of 0.1 M HCl, suppose H2S dissociates y M.

$\underset{\begin{array}{l}\\ Initial\text{conc}\text{. 0}\text{.1}\\ \\ \text{At equilibrium 0}\text{.1}-y\text{}\end{array}}{\left[{H}_{2}S\right]}\underset{}{⇌}\underset{\begin{array}{l}\\ \text{0}\\ \\ \text{y}\end{array}}{H{S}^{-}}+\underset{\begin{array}{l}\\ \text{0}\\ \\ \text{y}\end{array}}{{H}^{+}}$

Thus, at equilibrium,

[H2S] = 0.1 – y» 0.1,

[H+]= 0.1 + y » 0.1

[HS]= y M

${K}_{a}=\frac{0.1×y}{0.1}$

= 9.1 x 10-8

Thus, y = 9.1 x 10-8 M

(iii)The ionization of H2S is as follows –

$\underset{\begin{array}{l}\\ Initial\text{conc}\text{. (M) 9}{\text{.4×10}}^{-5}\text{}\\ \\ final\text{conc}\text{. (M) 9}{\text{.4×10}}^{-5}-X\end{array}}{{H}_{2}S}\underset{}{⇌}\underset{\begin{array}{l}\\ \text{9}{\text{.4×10}}^{-5}\\ \\ \text{X}\end{array}}{2{H}^{+}}+\underset{\begin{array}{l}\\ \text{0}\\ \\ \text{X}\end{array}}{{S}^{2-}}$

Therefore,

$\begin{array}{l}{K}_{{a}_{2}}=\frac{\left[{H}^{+}\right]\left[{S}^{2-}\right]}{\left[H{S}^{-}\right]}\\ {K}_{{a}_{2}}=\frac{\left(9.54×{10}^{-5}\right)\left(X\right)}{9.54×{10}^{-5}}\\ X=1.2×{10}^{-13}\end{array}$

Hence,

$\left[{S}^{2-}\right]=1.2×{10}^{-13}M$

In the presence of 0.1 M of HCl, let the concentration of S2- be Y M.

$\underset{\begin{array}{l}\\ Initial\text{conc}\text{. 0}\text{.1}\\ \\ final\text{conc}\text{. 0}\text{.1}-y\end{array}}{\left[{H}_{2}S\right]}\underset{}{⇌}\underset{\begin{array}{l}\\ \text{0}\\ \\ \text{y}\end{array}}{H{S}^{-}}+\underset{\begin{array}{l}\\ \text{0}\\ \\ \text{y}\end{array}}{{H}^{+}}$

And

$HCl\underset{}{⇌}\underset{\begin{array}{l}\\ \text{0}\text{.1M}\end{array}}{{H}^{+}}+\underset{\begin{array}{l}\\ 0.1M\end{array}}{C{l}^{-}}$ $\begin{array}{l}{K}_{{a}_{2}}=\frac{\left[{H}^{+}\right]\left[{S}^{2-}\right]}{\left[H{S}^{-}\right]}\\ 1.2×{10}^{-13}=\frac{\left(0.1\right)\left(Y\right)}{9.1×{10}^{-8}}\\ 10.92×{10}^{-21}=0.1Y\\ \frac{10.92×{10}^{-21}}{0.1}=Y\\ Y=1.092×{10}^{-19}M\\ \left[{S}^{2-}\right]=1.092×{10}^{-19}M\end{array}$

Q.33 Convert the following into basic units:

(i) 28.7 pm

(ii) 15.15 pm

(iii) 25365 mg

Ans.

(i) 28.7 pm:

1 pm = 10–12 m

∴ 28.7 pm = 28.7 × 10–12 m

= 2.87 × 10–11 m

(ii) 15.15 pm:

1 pm = 10–12 m

15.15 pm = 15.15 × 10–12 m

= 1.515 × 10–13 m

(iii) 25365 mg:

1 mg = 10–3 g

25365 mg = 2.5365 × 104 × 10–3 g

Since,

1 g = 10–3 kg

2.5365 × 101g = 2.5365 × 10–1 × 10–3 kg

∴ 25365 mg = 2.5365 × 10–2 kg

Q.34 Which one of the following will have largest number of atoms?

(i) 1 g Au (s)

(ii) 1 g Na (s)

(iii) 1 g Li (s)

(iv) 1 g of Cl2(g)

Ans.

$\begin{array}{l}1gofAu\left(s\right)=\frac{1}{197}molofAu\left(s\right)\\ =\frac{6.022×{10}^{23}}{197}atomsofAu\left(s\right)\\ =\text{3}.0\text{6}×\text{1}{0}^{\text{21}}\text{atoms of Au}\left(\text{s}\right)\\ 1gofNa\left(s\right)=\frac{1}{23}molofNa\left(s\right)\\ =\frac{6.022×{10}^{23}}{23}atomsofNa\left(s\right)\\ =\text{}0.\text{262}×\text{1}{0}^{\text{23}}\text{atoms of Na}\left(\text{s}\right)\\ =\text{26}.\text{2}×\text{1}{0}^{\text{21}}\text{atoms of Na}\left(\text{s}\right)\\ 1gofLi\left(s\right)=\frac{1}{7}molofLi\left(s\right)\\ =\frac{6.022×{10}^{23}}{7}atomsofLi\left(s\right)\\ =\text{}0.\text{86}×\text{1}{0}^{\text{23}}\text{atoms of Li}\left(\text{s}\right)\\ =\text{86}.0\text{}×\text{1}{0}^{\text{21}}\text{atoms of Li}\left(\text{s}\right)\\ 1gofC{I}_{2}\left(g\right)=\frac{1}{71}molofC{I}_{2}\left(g\right)\\ \left({\text{Molar mass of Cl}}_{\text{2}}\text{molecule}=\text{35}.\text{5}×\text{2}={\text{71 g mol}}^{-\text{1}}\right)\\ =\frac{6.022×{10}^{23}}{71}atomsofC{I}_{2}\left(g\right)\\ =\text{}0.0\text{848}×\text{1}{0}^{\text{23}}{\text{atoms of Cl}}_{\text{2}}\left(\text{g}\right)\\ =\text{8}.\text{48}×\text{1}{0}^{\text{21}}{\text{atoms of Cl}}_{\text{2}}\left(\text{g}\right)\end{array}$

Hence, 1 g of Li (s) will have the largest number of atoms.

Q.35 The ionization constant of acetic acid is 1.74 × 10–5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the conc. of acetate ion in the solution and its pH.

Ans.

The given reaction is

$\begin{array}{l}C{H}_{3}COOH\underset{}{⇌}C{H}_{3}CO{O}^{-}+{H}^{+}\\ \text{Degree of dissociation},\\ \alpha =\sqrt{{K}_{a}/C}\end{array}$

Ka= ionization constant

C = concentration

C = 0.05 M Ka= 1.74 × 10–5

$\begin{array}{l}\alpha =\sqrt{\frac{1.74×{10}^{-5}}{5×{10}^{-2}}}\\ =\sqrt{\left(34.8×{10}^{-5}\right)}\end{array}$

= 1.86 x 10-2

[CH3COO]= = 0.05 x 1.86 x 10-2

= 0.093 x 10-2 M

[CH3COO] = [H+] = 0.093 x 10-2 M

pH = – log [H+]= – log (0.093 x 10-2)

or, pH = 3.03

Q.36 It has been found that the pH of a 0.01M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pKa.

Ans.

Let the organic acid be HA.

$HA\underset{}{⇌}{H}^{+}+{A}^{-}$

Concentration of HA = 0.01 M

pH = 4.15

$\begin{array}{l}⇒-\mathrm{log}\left[{H}^{+}\right]=4.15\\ ⇒\left[{H}^{+}\right]=anti\mathrm{log}\left(-4.15\right)\\ ⇒\left[{H}^{+}\right]=7.08×{10}^{-5}\end{array}$

Let us calculate ionization constant by the expression mentioned here.

$\begin{array}{l}{K}_{a}=\frac{\left[{H}^{+}\right]\left[{A}^{-}\right]}{\left[HA\right]}\\ \left[{H}^{+}\right]=\left[{A}^{-1}\right]=7.08×{10}^{-5}\\ \left[{H}^{+}\right]=0.01\end{array}$

Then,

${K}_{a}=\frac{\left(7.08×{10}^{-5}\right)\left(7.08×{10}^{-5}\right)}{\left(0.01\right)}$

⇒Ka= 5.01 x 10-7

⇒pKa = – log Ka

= – log (5.01 x 10-7)

= 6.3001

Q.37 Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040 (assume the density of water to be one).

Ans.

Q.38 What will be the mass of one 12C atom in g?

Ans.

1 mole of carbon atoms = 6.023 × 1023 atoms of carbon

= 12 g of carbon

$\therefore {\text{Mass of one}}^{\text{12}}\text{C atom}=\frac{12g}{6.022×{10}^{23}}$

= 1.993 × 10–23g

Q.39 Assuming complete dissociation, calculate the pH of the following solutions:

(a) 0.003 M HCl (b) 0.005 M NaOH (c) 0.002 M HBr (d) 0.002 M KOH

Ans.

(a) 0.003 M HCl:

Considering the equilibrium: <

$HCl\left(aq\right)\underset{}{⇌}{H}^{+}\left(aq\right)+C{l}^{-}\left(aq\right)$

Thus, [H+] = [HCl] = 3 x 10–3

pH= – log [H+]= –log (3 x 10–3)= 2.52

(b) 0.005 M NaOH:

Considering the equilibrium:

$NaOH\left(aq\right)\underset{}{⇌}N{a}^{+}\left(aq\right)+O{H}^{-}\left(aq\right)$

Thus, [OH] = 5 x 10–3 M

From the relation,

[H+] [OH] = 1 x 10-14

Or, [H+] = (1 x 10–14) / (5 x 10–3)

= 2 x 10–12

pH= – log [H+] = – log (2 x 10–12) = 11.70

(c) 0.002 M HBr:

Considering the equilibrium:

$HBr\left(aq\right)\underset{}{⇌}{H}^{+}\left(aq\right)+B{r}^{-}\left(aq\right)$

[H+] = 2 x10–3 M

pH = – log [H+] = – log (2 x 10–3) = 2.70

(d) 0.002 M KOH:

Considering the equilibrium:

$KOH\left(aq\right)\underset{}{⇌}{K}^{+}\left(aq\right)+O{H}^{-}\left(aq\right)$

Thus, [OH] = 2 x 10–3 M

From the relation,

[H+] [OH] = 1 x 10-14

or [H+] = (1 x 10–14) / (2 x 10–3) = 5 x 10–12

pH = – log [H+] = – log (5 x 10–12) = 11.30

Q.40 How many significant figures should be present in the answer of the following calculations:

$\left(\mathrm{i}\right)\frac{0.02856×298.15×0.112}{0.5785}$

(ii) 5 X 5.364

(iii) 0.0125 X 0.78640 X.0215

Ans.

Least precise number of calculation = 0.112

Number of significant figures in the answer

= Number of significant figures in the least precise number

= 3

(ii) 5 × 5.364

Least precise number of calculation = 5.364

Number of significant figures in the answer = Number of significant figures in 5.364

= 4

(iii) 0.0125 + 0.7864 + 0.0215

Since the least number of decimal places in each term is four, the number of significant

figures in the answer is also 4.

Q.41 Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes:

 Isotope Isotopic molar mass Abundance 36Ar 35.96755 gmol–1 0.337% 38Ar 37.96272 gmol–1 0.063% 40Ar 39.9624 gmol–1 99.600%

Ans.

Molar mass of argon

$\begin{array}{l}\left[\left(35.96755×\frac{0.337}{100}\right)+\left(37.96272×\frac{0.063}{100}\right)+\left(39.9624×\frac{90.60}{100}\right)\right]gmo{l}^{-1}\\ =\left[0.121+0.024+39.802\right]gmo{l}^{-1}\end{array}$

= 39.947gmol–1

Q.42 Calculate the pH of the following solutions:

1. 2 g of TlOH dissolved in water to give 2 litre of the solution.
2. 0.3 g of Ca(OH)2 dissolved in water to give 500 mL of the solution.
3. 0.3 g of NaOH dissolved in water to give 200 mL of the solution
4. 1 mL of 13.6 M HCl is diluted with water to give 1 litre of the solution.

Ans.

(a)

$\begin{array}{l}Molar\text{conc}\text{. of TIOH=}\frac{2g}{221gmo{l}^{-1}}×\frac{1}{2L}\\ =4.52×{10}^{-3}\end{array}$

[OH]= [TlOH] = 4.52 x 10–3 M

[H+] [OH] = 1 x 10-14

[H+]= (1 x 10–14)/ (4.52 x 10–3) = 2.21 x 10–12 M

pH = – log (2.21 x 10–12) = 12 – (0.3424) = 11.66

(b)

$\begin{array}{l}Molar\text{conc}\text{. of Ca}{\left(OH\right)}_{2}\text{=}\frac{0.3g}{74gmo{l}^{-1}}×\frac{1}{0.5\text{}L}\\ =8.11×{10}^{-3}\\ Ca{\left(OH\right)}_{2}\left(aq\right)\underset{}{⇌}C{a}^{2+}\left(aq\right)+2O{H}^{-}\left(aq\right)\end{array}$

[OH] = 2 [Ca(OH)2]

= 2 x (8.11 x 10–3) M

= 16.22 x 10–3 M

pOH = – log (16.22 x 10–3)

= 3 – 1.2101

= 1.79

pH = 14 – 1.79

(c)

$\begin{array}{l}Molar\text{conc}\text{. of NaOH=}\frac{0.3g}{40\text{}gmo{l}^{-1}}×\frac{1}{0.2\text{}L}\\ =3.75×{10}^{-2}M\end{array}$

[OH]= [NaOH] = 3.75 x 10–2 M

pOH = – log (3.75 x 10–2 M) = 2 – 0.0574 = 1.43

pH = 14 – 1.43 = 12.57

(d) From the relation,

M1V1 = M2V2

or 13.6 M x 1 mL = M2 x 1000 mL

or M2 = 1.36 x 10–2 M

[H+] = [OH] = 1.36 x 10–2 M

pH = – log (1.36 x 10–2) = 2 – 0.1335 = 1.87

Q.43 Calculate the number of atoms in each of the following (i) 52 moles of Ar (ii) 52 u of He (iii) 52 g of He.

Ans.

(i) 1 mole of Ar = 6.022 × 1023 atoms of Ar
∴ 52 mol of Ar = 52 × 6.022 × 1023 atoms of Ar
= 3.131 × 1025 atoms of Ar
(ii) 1 atom of He = 4 u of He
Or,
4 u of He = 1 atom of He

$\begin{array}{l}1uofHe=\frac{1}{4}atomofHe\\ 52uofHe=\frac{52}{4}atomofHe\end{array}$

= 13 atoms of He
(iii) 4 g of He = 6.022 × 1023 atoms of He

$52gofHe=\frac{6.022×{10}^{23}×52}{4}atomofHe$

= 7.8286 × 1024 atoms of He

Q.44 The degree of ionization of a 0.1 M bromo acetic acid solution is 0.132. Calculate the pH of the solution and the pKa of bromoacetic acid.

Ans.

Degree of ionization, a = 0.132

Concentration, c = 0.1 M

Thus, the concentration of H3O+= c.a = 0.1 × 0.132

= 0.0132

$\begin{array}{l}pH=-\mathrm{log}\left[{H}^{+}\right]\\ \text{}=\text{}–\text{log}\left(0.0\text{132}\right)\text{}=\text{1}.\text{88}\\ {\text{K}}_{a}=C{\alpha }^{2}\end{array}$

= 0.1 x (0.132)2

Ka = 0.0017

and pKa = – log (0.0017)

= 2.77

Q.45 A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate (i) empirical formula, (ii) molar mass of the gas, and (iii) molecular formula.

Ans.

$\begin{array}{l}\left(\text{i}\right)\text{1 mole}\left(\text{44 g}\right){\text{of CO}}_{\text{2}}\text{contains 12 g of carbon}.\\ \therefore 3.38gofC{O}_{2}willcontaincarbon=\frac{12g}{44g}×3.38g\\ =\text{}0.\text{9217 g}\\ \text{18 g of water contains 2 g of hydrogen}.\\ \therefore 0.690gofwaterwillcontainhydrogen=\frac{2g}{18g}×0.690g\\ =\text{}0.0\text{767 g}\\ \text{Since carbon and hydrogen are the only constituents of the compound},\text{the total mass of the compound is}:\\ =\text{}0.\text{9217 g}+\text{}0.0\text{767 g}\\ =\text{}0.\text{9984 g}\\ percentofCinthecompound=\frac{0.9217g}{0.9984g}×100\\ =\text{92}.\text{32}%\\ percentofHinthecompound=\frac{0.0767g}{0.9984g}×100\\ =\text{7}.\text{68}%\\ Molesofcarboninthecompound=\frac{92.32}{12.00}\\ =\text{7}.\text{69}\\ Molesofhydrogeninthecompound=\frac{7.68}{1}\\ =\text{7}.\text{68}\\ \text{Ratio of carbon to hydrogen in the compound}=\text{7}.\text{69}:\text{7}.\text{68}\\ =\text{1}:\text{1}\\ \text{Hence},\text{the empirical formula of the gas is CH}.\\ \left(\text{ii}\right)\text{Given},\\ \text{Weight of 1}0.0\text{L of the gas}\left(\text{at S}.\text{T}.\text{P}\right)\text{}=\text{11}.\text{6}\\ \therefore weightof22.4LofgasatSTP=\frac{11.6g}{10.0L}×22.4L\\ =\text{25}.\text{984 g}\\ \approx \text{26 g}\\ \text{Hence},\text{the molar mass of the gas is 26 g}.\\ \left(\text{iii}\right)\text{Empirical formula mass of CH}=\text{12}+\text{1}=\text{13 g}\\ n=\frac{Molarmassofgas}{Empiricalformulamassofgas}\\ =\frac{26g}{13g}\\ \text{n}=\text{2}\\ \text{Molecular formula of gas}=\text{}{\left(\text{CH}\right)}_{\text{n}}\\ ={\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\end{array}$

Q.46 The pH of 0.005M codeine (C18H21NO3) solution is 9.95. Calculate the ionization constant and pKb.

Ans.

Considering the reaction

$Cod+{H}_{2}O\underset{}{⇌}Cod{H}^{+}+O{H}^{-}$

pH = 9.95, pOH = 14 – pH

= 14 – 9.95

= 4.05

or – log [OH] = 4.05

or log[OH] = – 4.05

or [OH] = antilog (– 4.05) = 8.913 x 10-5

$\begin{array}{l}c\alpha =8.91×{10}^{-5}\\ \alpha =\frac{8.91×{10}^{-5}}{5×{10}^{-3}}=1.782×{10}^{-2}\\ {K}_{b}=c{\alpha }^{2}\\ {K}_{b}=0.005×{\left(1.782\right)}^{2}×{10}^{-4}\\ \text{}=0.005×3.1755×{10}^{-4}\\ \text{}=0.005×{10}^{-4}\\ \text{}=1.58×{10}^{-6}\end{array}$

pKb = – log (1.588 x 10–6) = 6 – 0.1987 = 5.8

Q.47 Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction, CaCO3(s) + 2HCl(aq) → CaCl2(aq)+ CO2(g) + H2O(l)

What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl?

Ans.

0.75 M of HCl ≡ 0.75 mol of HCl are present in 1 L of water
≡ [(0.75 mol) × (36.5 g mol–1)] HCl is present in 1 L of water
≡ 27.375 g of HCl is present in 1 L of water
Thus, 1000 mL of solution contains 27.375 g of HCl.
Amount of HCl present in 25 mL of solution

$=\frac{27.375g}{1000mL}×25mL$

= 0.6844 g
From the given chemical equation,
CaCO3(s)+ 2HCl(aq)→CaCl2(aq)+CO2(g)+H2O(l)
2 mol of HCl (2 × 36.5 = 73 g) react with 1 mol of CaCO3 (100 g).

${\text{Amount of CaCO}}_{\text{3}}\text{that react with}0.6844g=\frac{100}{73}×0.6844g$

=0.938 g

Q.48 Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous hydrochloric acid according to the reaction

4HCl(aq) + MnO2(s) → 2H2O(l) + MnCl2(aq) + Cl2(g)

How many grams of HCl react with 5.0 g of manganese dioxide?

Ans.

1 mol [55 + 2 × 16 = 87 g] MnO2 reacts completely with 4 mol

[4 × 36.5 = 146 g] of HCl.

5.0 g of MnO2 will react with

$=\frac{146g}{87g}×5.0gofHCL$

= 8.4 g of HCl
Hence, 8.4 g of HCl will react completely with 5.0 g of manganese dioxide.

Q.49 What is the pH of 0.001 M aniline solution? The ionization constant of aniline = 4.27 x 10-10. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.

Ans.

Kb= 4.27 × 10–10 and c = 0.001 M

Kb= cα2

$\begin{array}{l}⇒4.27×{10}^{-10}=0.001×{\alpha }^{2}\\ ⇒4270×{10}^{-10}={\alpha }^{2}\\ ⇒65.34×{10}^{-5}=\alpha =6.53×{10}^{-4}\end{array}$

Thus, [OH] = c . α = 0.001 x 65.34 x 10–5

= 0.065 x 10–5

pOH = – log [OH]

= – log (0.065 x 10–5)

= – log (65 x 10–8)

= 6.187

We know,

pH + pOH = 14

or pH = 14 – 6.187 = 7.813

Again,

Ka x Kb = Kw

or Ka x (4.27 × 10–10) = 10–14

${K}_{a}=\frac{{10}^{-14}}{4.27×{10}^{-10}}=2.34×{10}^{-5}$

The ionization constant of the conjugate acid of aniline is 2.34 × 10–5.

Q.50 Calculate the degree of ionization of 0.05 M acetic acid if its pKa value is 4.74. How is the degree of dissociation affected when its solution also contains (a) 0.01 M (b) 0.1 M in HCl?

Ans.

Given values are as follows-

C = 0.05 M = 5 x 10-2

pKa = 4.74

pKa = – log Ka = 4.74

or Ka = antilog (– 4.74) = 1.82 x 10–5

From the relation,

Ka = cα2

$\begin{array}{l}⇒\alpha =\sqrt{\left({K}_{a}/c\right)}\\ ⇒\alpha =\sqrt{\frac{1.82×{10}^{-5}}{5×{10}^{-2}}}\end{array}$

= 1.908 x 10–2

When HCl is added to the solution, the concentration of H+ ions increases. Thus the equilibrium will shift in the backward direction. As a result, the dissociation of acetic acid will decrease.

1. When 0.01 M HCl is taken:

Let x be the amount of acetic acid dissociated after the addition of HCl.

$\underset{\begin{array}{l}\\ Initial\text{conc}\text{. 0}\text{.05 M}\\ \\ \text{At equilibrium}\left(0.05-x\right)\end{array}}{C{H}_{3}COOH}\underset{}{⇌}\underset{\begin{array}{l}\\ 0\\ \\ \left(0.01+x\right)\end{array}}{{H}^{+}}+\underset{\begin{array}{l}\\ 0\\ \\ X\end{array}}{C{H}_{3}COO-}$

As a very small amount of acetic acid is dissociated, the values of (0.05 – x) and (0.01 + x) can be taken as 0.05 M and 0.01 M respectively.

$\begin{array}{l}{K}_{a}=\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[{H}^{+}\right]}{\left[C{H}_{3}COOH\right]}\\ \text{Putting the values in the above equation}:\\ {K}_{a}=\frac{\left(0.01\right)x}{0.05}\\ ⇒x=\frac{1.82×{10}^{-5}×0.05}{0.01}\\ ⇒x=1.82×{10}^{-3}×0.05\text{}M\\ \text{By definition},\\ \alpha =\frac{Amount\text{of aicd dissociated}}{Amount\text{of acid taken}}\\ ⇒\alpha =\frac{1.82×{10}^{-3}×0.05}{0.05}\end{array}$

=1.82 x 10–3
1. When 0.1 M HCl is taken.
Let us assume that the amount of acetic acid dissociated in this case is X. The concentrations of various species involved in the reaction are:
[CH3COOH]= 0.05 – X ~ 0.05M
[CH3COO] = X
[H+]= 0.1 + X ~ 0.1

$\begin{array}{l}{K}_{a}=\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[{H}^{+}\right]}{\left[C{H}_{3}COOH\right]}\\ {K}_{a}=\frac{\left(0.01\right)x}{0.05}\\ ⇒x=\frac{1.82×{10}^{-5}×0.05}{0.01}\\ ⇒x=1.82×{10}^{-3}×0.05\text{}M\\ Now,\\ \alpha =\frac{Amount\text{of aicd dissociated}}{Amount\text{of acid taken}}\\ ⇒\alpha =\frac{1.82×{10}^{-4}×0.05}{0.05}\\ =\text{1}.\text{82 x 1}{0}^{–\text{4}}\end{array}$

Q.51 What will be the minimum pressure required to compress 500 dm3 of air at 1 bar to 200dm3 at 30°C?

Ans.

Given,

Initial pressure, p1 = 1 bar

Initial volume, V1 = 500 dm3

Final volume, V2 = 200 dm3

Since the temperature remains constant, the final pressure (p2) can be calculated using Boyle’s law.

According to Boyle’s law,

$\begin{array}{l}{p}_{1}{v}_{1}={p}_{2}{v}_{2}\\ Or,\text{}{p}_{2}=\frac{{p}_{1}{v}_{1}}{{v}_{2}}\\ =\frac{1×500}{200}bar\\ =2.5\text{bar}\end{array}$

Therefore, the minimum pressure required is 2.5 bar.

Q.52 A vessel of 120 mL capacity contains a certain amount of gas at 35 °C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35 °C. What would be its pressure?

Ans.

Given,

Initial pressure, p1 = 1.2 bar

Initial volume, V1 = 120 mL

Final volume, V2 = 180 mL

Since the temperature remains constant, the final pressure (p2) can be calculated using Boyle’s law.

According to Boyle’s law,

$\begin{array}{l}{p}_{1}{v}_{1}={p}_{2}{v}_{2}\\ {p}_{2}=\frac{{p}_{1}{v}_{1}}{{v}_{2}}\\ =\frac{1.2×120}{180}bar\\ =0.8\text{bar}\end{array}$

Therefore, the pressure would be 0.8 bar.

Q.53 Using the equation of state pV = nRT; show that at a given temperature density of a gas is proportional to gas pressure p.

Ans.

The equation of state is given by,

pV = nRT ……….. (i)

Where,

p →Pressure of gas

V →Volume of gas

n→ Number of moles of gas

R→ Gas constant

T →Temperature of gas

From equation (i) we have,

$\begin{array}{l}\frac{n}{V}=\frac{p}{RT}\text{}\dots \dots \dots \dots \left(ii\right)\\ \mathrm{Re}placing\text{}n\text{with}\frac{m}{M}\text{in equation (ii), we have}\\ \frac{m}{MV}=\frac{p}{RT}\text{}\dots \dots \dots \dots \left(iii\right)\end{array}$

Where,

m →Mass of gas

M →Molar mass of gas

$\begin{array}{l}But\text{}\frac{m}{MV}=d\left(d=density\text{of gas}\right)\\ \text{from equation}\left(\text{iii}\right)\\ \frac{d}{M}=\frac{p}{RT}\\ ⇒d=\left(\frac{M}{RT}\right)p\end{array}$

Molar mass (M) of a gas is always constant so, at constant temperature (T)

$\begin{array}{l}\frac{M}{RT}=cons\mathrm{tan}t\\ d=\left(cons\mathrm{tan}t\right)p\\ ⇒d\propto p\end{array}$

So, at a given temperature, the density (d) of gas is proportional to its pressure (p)

Q.54 At 0°C, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?

Ans.

Density (d) of the substance at temperature (T) can be given as,

$\begin{array}{l}d=\frac{M}{RT}\\ \text{Density of oxide}\left({\text{d}}_{\text{1}}\right)\text{is given by},\\ {d}_{1}=\frac{{M}_{1}{p}_{1}}{RT}\end{array}$

Here, M1 and p1 are the mass and pressure of the oxide respectively.
Density of dinitrogen gas (d2) can be given as:

${d}_{2}=\frac{{M}_{2}{p}_{2}}{RT}$

In above equation, M2 and p2 are the mass and pressure of the oxide respectively.
According to the given question,

$\begin{array}{l}{d}_{1}={d}_{2}\\ \therefore {M}_{1}{p}_{1}={M}_{2}{p}_{2}\end{array}$

Molecular mass of nitrogen, M = 28 g/mol

$\begin{array}{l}Now,{\text{M}}_{1}=\frac{{M}_{2}{P}_{2}}{{p}_{1}}\\ \frac{28×5}{2}\end{array}$

=70 g/mol

Therefore, the molecular mass of the oxide is 70 g/mol.

Q.55 Pressure of 1 g of an ideal gas A at 27 °C is found to be 2bar. When 2 g of another ideal gas B is introduced in the same flask at same temperature the pressure becomes 3bar.Find a relationship between their molecular masses.

Ans.

The ideal gas equation for ideal gas A is given by,

pAV = nART

Here, pA and nA is the pressure and number of moles of gas A.

For ideal gas B, the ideal gas equation is:

${p}_{B}V={n}_{B}RT\text{}\dots \dots \dots \dots .\left(ii\right)$

Where, p and n represent the pressure and number of moles of gas B.

[V and T are constants for gases A and B]

From equation (i),

$\begin{array}{l}{p}_{A}V=\frac{{m}_{A}}{{M}_{A}}RT\\ ⇒\frac{{p}_{A}{M}_{A}}{{m}_{A}}=\frac{RT}{V}\text{}\text{}\dots \dots \dots \left(iii\right)\end{array}$

From equation (ii),

$\begin{array}{l}{p}_{B}V=\frac{{m}_{B}}{{M}_{B}}RT\\ ⇒\frac{{p}_{B}{M}_{B}}{{m}_{B}}=\frac{RT}{V}\text{}\text{}\dots \dots \dots \left(iv\right)\end{array}$

Here, MA and MB are the molecular masses of gases A and B respectively.

From equations (iii) and (iv),

$\begin{array}{l}{p}_{B}V=\frac{{m}_{B}}{{M}_{B}}RT\\ ⇒\frac{{p}_{B}{M}_{B}}{{m}_{B}}=\frac{RT}{V}\text{}\text{}\dots \dots \dots \left(iv\right)\\ Given,\\ {m}_{A}=1g\\ {p}_{A}=2\text{bar}\\ {m}_{B}=2g\\ {p}_{B}=\left(3-2\right)=1\text{bar}\end{array}$

(Since total pressure is 3 bar)

Substituting the above values in equation (v)

$\begin{array}{l}\frac{2×{M}_{A}}{1}=\frac{1×{M}_{B}}{2}\\ ⇒4{M}_{A}={M}_{B}\\ \text{Thus},\text{the relationship of molecular masses of A and B can be given as}:\\ 4{M}_{A}={M}_{B}\end{array}$

Q.56 A thermodynamic state function is a quantity

(i) Used to determine heat changes

(ii) Whose value is independent of path

(iii) Used to determine pressure volume work

(iv) Whose value depends on temperature only.

Ans.

Explanation:

A thermodynamic state function is a quantity whose value is independent of a path.

Functions like p, V, T etc. depend only on the state of a system and not on the path.

Q.57 For the process to occur under adiabatic conditions, the correct condition is:

(i) ∆T = 0

(ii) ∆p = 0

(iii) q = 0

(iv) w = 0

Ans.

Explanation:

When there is no exchange of heat between the system and its surroundings then the system is said to be under adiabatic conditions. Hence, under adiabatic conditions, q = 0.

Q.58 The enthalpies of all elements in their standard states are:

(i) unity

(ii) zero

(iii) < 0

(iv) Different for each element

Ans.

Explanation:

The enthalpy of all elements in their standard state is zero.

Q.59 ∆Uθof combustion of methane is – X kJ mol–1. The value of ∆Hθ is

(i) = ∆Uθ

(ii) > ∆Uθ

(iii) < ∆Uθ

(iv) = 0

Ans.

Explanation:

Since ∆Hθ = ∆Uθ + ∆ngRT and ∆Uθ = –X kJ mol–1,

∆Hθ = (–X) + ∆ngRT.

⇒ ∆Hθ < ∆Uθ

Q.60 The drain cleaner, Drainex contains small bits of aluminum which react with caustic soda to produce dihydrogen. What volume of dihydrogen at 20 °C and one bar will be released when 0.15g of aluminum reacts?

Ans.

The reaction of aluminium with caustic soda (NaOH)can be represented as:

$\underset{\begin{array}{l}\\ 2×27g\end{array}}{2Al}+2NaOH+2{H}_{2}O\to 2NaAI{O}_{2}+\underset{\begin{array}{l}\\ 3×22400mL\end{array}}{3{H}_{2}}$

At STP (273.15 K and 1 atm),

Hence, 54 g (2 × 27 g) of aluminium gives 3 × 22400 mL of H2.

$\therefore 0.15gAl\text{gives}\frac{3×22400×0.15}{54}mL{\text{of H}}_{2}\text{}i.e.,186.67{\text{mL of H}}_{2}$

At STP,

$\begin{array}{l}{p}_{1}=1\text{}atm\\ {V}_{1}=186.67\text{mL}\\ {\text{T}}_{1}=273.15\text{}K\end{array}$

Let the volume of dihydrogen be V2 at p2 = 0.987 atm (since 1 bar = 0.987 atm) and T2

= 20°C = (273.15 + 20) K = 293.15 K

$\begin{array}{l}\frac{{p}_{1}{V}_{1}}{{T}_{1}}=\frac{{p}_{2}{V}_{2}}{{T}_{2}}\\ ⇒{V}_{2}=\frac{{p}_{1}{V}_{1}{T}_{2}}{{p}_{2}{T}_{1}}\\ =\frac{1×186.67×293.15}{0.987×273.15}\\ =202.98\text{mL}\\ \text{=203 mL}\end{array}$

Therefore, 203 mL of dihydrogen will be released.

Q.61 The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, –890.3 kJ mol–1 ,–393.5 kJ mol–1, and –285.8 kJ mol–1 respectively. Enthalpy of formation of CH4 (g) will be

(i) –74.8 kJ mol–1 (ii) –52.27 kJ mol–1

(iii) +74.8 kJ mol–1 (iv) +52.26 kJ mol–1.

Ans.

Explanation:

Given: The enthalpy of combustion of,

1. CH4(g) + 2O2(g) → CO2(g) +2H2O(g) ∆H=-890.3 kJ mol -1
2. C(s)+O2(g) → CO2(g) ∆H=-393.5kJ mol -1
3. 2H2(g)+ O2(g) → 2H2O(g) ∆H=-285.8 kJ mol -1

Thus, the desired equation is the one that represents the formation of CH4

C(s)+2H2(g) → CH4(g)

fHCH4 =∆cHc+2∆cHH2 -∆CHCO2

=[-393.5+2(-285.8)-(-890.3)]kJ mol-1

=-74.8 kJ mol-1

Enthalpy of formation of CH4(g) = –74.8 kJ mol–1

Hence, alternative (i) is correct.

Q.62 The ionization constant of dimethylamine is 5.4 × 10–4. Calculate its degree of ionization in its 0.02 M solution. What percentage of dimethylamine is ionized if the solution is also 0.1 M in NaOH?

Ans.

c = 0.02 M and Kb = 5.4 × 10–4

$\begin{array}{l}\alpha =\sqrt{\left({K}_{b}/c\right)}\\ =\sqrt{\frac{5.4×{10}^{-4}}{0.02}}\end{array}$

=0.1643

When 0.1 M of NaOH is added to the solution, then NaOH (being a strong base) undergoes complete ionization.

$\begin{array}{l}NaOH\left(aq\right)\underset{}{⇌}\underset{\begin{array}{l}\\ 0.1\text{M}\end{array}}{N{a}^{+}\left(aq\right)}+\underset{\begin{array}{l}\\ 0.1\text{M}\end{array}}{O{H}^{-}\left(aq\right)}\\ \underset{\left(0.02-x\right)~0.02}{\underset{}{{\left(C{H}_{3}\right)}_{2}NH}}+{H}_{2}O\underset{}{⇌}\underset{\begin{array}{l}\\ x\end{array}}{{\left(C{H}_{3}\right)}_{2}N{H}_{2}^{+}}+\underset{\begin{array}{l}\\ \left(x+0.1\right)~0.1\end{array}}{O{H}^{-}}\\ \left[{\left(C{H}_{3}\right)}_{2}N{H}_{2}^{+}\right]=x\\ \left[{\text{OH}}^{–}\right]\text{}=\text{x}+\text{}0.\text{1}~\text{}0.\text{1}\\ {\text{K}}_{b}=\frac{\left[{\left(C{H}_{3}\right)}_{2}N{H}_{2}^{+}\right]\left[O{H}^{-}\right]}{{\left(C{H}_{3}\right)}_{2}NH}\\ ⇒5.4×{10}^{-4}=\frac{x×0.1}{0.02}\end{array}$

⇒x = 0.0054

It means that in the presence of 0.1 M NaOH, 0.54% of dimethylamine will get dissociated.

Q.63 What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a 9 dm flask at 27 °C?

Ans.

Since,

$\begin{array}{l}p=\frac{m}{M}\frac{RT}{V}\\ \text{For methane}\left({\text{CH}}_{\text{4}}\right),\\ {p}_{C{H}_{4}}=\frac{3.2}{16}×\frac{8.314×300}{9×{10}^{-3}}\left[\frac{Since\text{}9d{m}^{3}=9×{10}^{-3}{m}^{3}}{27ºC=300\text{}K}\right]\\ =5.543×{10}^{4}pa\\ \text{For carbon dioxide}\left({\text{CO}}_{\text{2}}\right),\\ {p}_{C{O}_{2}}=\frac{4.4}{44}×\frac{8.314×300}{9×{10}^{-3}}\\ =2.771×{10}^{4}pa\\ \text{Total pressure exerted by the mixture can be obtained as}:\\ p={p}_{C{H}_{4}}+{p}_{C{O}_{2}}\\ =\left(5.543×{10}^{4}+2.771×{10}^{4}\right)pa\\ =8.314×{10}^{4}pa\end{array}$

Therefore, total pressure exerted by the mixture is 8.314 × 104 Pa.

Q.64 A reaction, A + B → C + D + q is found to have a positive entropy change. The reaction will be

(i) possible at high temperature

(ii) possible only at low temperature

(iii) not possible at any temperature

(iv) possible at any temperature

Ans.

For a reaction to be spontaneous, ∆G should be negative.

∆G = ∆H – T∆S

According to the question, for the given reaction,

∆S = positive

∆H = negative (since heat is evolved)

⇒ ∆G = negative

Therefore, the reaction is spontaneous at any temperature.

Hence, alternative (iv) is correct.

Q.65 In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?

Ans.

According to the first law of thermodynamics,

∆U = q + W (i)

Where,

∆U = change in internal energy for a process

q = heat

W = work

Given,

q = + 701 J (Since heat is absorbed)

W = –394 J (Since work is done by the system)

Substituting the values in expression (i), we get

∆U = 701 J + (–394 J)

∆U = 307 J

Hence, the change in internal energy for the given process is 307 J.

Q.66 Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below:

(a) Human muscle-fluid, 6.83

(b) Human stomach-fluid, 1.2

(c) Human blood, 7.38

(d) Human saliva, 6.4.

Ans.

(a) For Human muscle fluid,

pH = 6.83

or pH = – log [H+]

or 6.83 = – log [H+]

or [H+] = antilog (-6.83)

or [H+] = 1.48 × 10–7 M

(b) For Human stomach fluid,

pH = 1.2 = – log [H+]

or [H+] = antilog (-1.2)

or [H+] = 0.063 M

(c) For Human blood,

pH = 7.38 = – log [H+]

or [H+] = antilog (-7.38)

or [H+] = 4.17 × 10–8 M

(d) For Human saliva,

pH = 6.4 = – log [H+]

or [H+] = antilog (-6.4)

or [H+] = 3.98 × 10–7 M

Q.67 The reaction of cyanamide, NH2CN(s), with dioxygen was carried out in a bomb calorimeter, and ∆U was found to be –742.7 kJ mol–1 at298 K. Calculate enthalpy change for the reaction at 298 K.

NH2CN(g)+ O2(g) → N2(g) +CO2(g) +H2O(l)

Ans.

Enthalpy change for a reaction (∆H) is given by the expression,

∆H = ∆U + ∆ngRT

Where,

∆U = change in internal energy

∆ng = change in number of moles

For the given reaction,

∆ng = ∑ng (products) – ∑ng (reactants)

= (2 – 1.5) moles

∆ng = 0.5 moles

And,

∆U = –742.7 kJ mol–1

T = 298 K

R = 8.314 × 10–3 kJ mol–1 K–1

Substituting the values in the expression of ∆H:

∆H = (–742.7 kJ mol–1) + (0.5 mol) (298 K) (8.314 × 10–3 kJ mol–1 K–1)

= 742.7 – 1.2

∆H = 741.5 kJ mol–1

Q.68 The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion concentration in each.

Ans.

The hydrogen ion concentration can be easily calculated by the following relation

$pH=-\mathrm{log}\left[{H}^{+}\right]$

1. pH of milk = 6.8

Using the above relation,

pH = – log [H+]

⇒ log [H+] = –6.8

⇒ [H+] = antilog (–6.8)

= 1.5 x 10–7 M

1. pH of black coffee = 5.0

Since, pH = – log [H+]

⇒ 5.0 = – log [H+]

⇒ log [H+] = –5.0

⇒ [H+] = antilog(–5.0) = 10–5 M

1. pH of tomato juice = 4.2

pH = – log [H+]

⇒ 4.2 = – log [H+]

⇒ log [H+] = –4.2

⇒ [H+] = antilog(–4.2)

=6.31 x10–5 M

1. pH of lemon juice = 2.2

pH = – log [H+]

⇒ 2.2 = – log [H+]

⇒ log [H+] = –2.2

⇒[H+] = antilog(–2.2)

=6.31 x10–3 M

1. pH of egg white = 7.

pH = – log [H+]

⇒7.8 = – log [H+]

⇒log [H+] = –7.8

⇒ [H+] = antilog(–7.8)

=1.58 x 10–8 M

Q.69 If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH?

Ans.

Let us calculate the concentration of KOH in the solution:

$\begin{array}{l}{\left[KOH\right]}_{\left(aq\right)}=\frac{0.561}{200/1000}g/L\\ =2.805g/L\\ =\frac{2.805}{56.11}M\\ =0.05\text{M}\\ \text{The ionic equilibrium for the aqueous KOH is given below}:\\ KOH\left(aq\right)\underset{}{⇌}{K}^{+}\left(aq\right)+O{H}^{-}\left(aq\right)\end{array}$

Thus, [K+] = [OH] = 0.05M
[H+] x [OH] = Kw
or [H+] = Kw/ [OH]

$=\frac{{10}^{-14}}{0.05}=2×{10}^{-13}\text{M}$

pH = – log [H+]
⇒ pH = – log (2 x 10–13) = 12.70

Q.70 Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol–1 K–1.

Ans.

From the expression of heat (q),

q = m. c. ∆T

Where,

c = molar heat capacity

m = mass of substance

∆T = change in temperature

Substituting the values in the expression of q:

q=[60/27 mole](24 j mole-1 K-1)(20K)

q = 1066.7 J

q = 1.07 kJ

Q.71 The solubility of Sr(OH)2 at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.

Ans.

Molecular mass of Sr(OH)2 = 87.6 + 34 = 121.6 gmol-1

Concentration of Sr(OH)2

$\begin{array}{l}=\frac{19.23g{L}^{-1}}{121.6gmo{l}^{-1}}\\ =0.1581\text{M}\\ \text{Assuming complete dissociation of Sr}{\left(\text{OH}\right)}_{\text{2}}\\ Sr{\left(OH\right)}_{2}\underset{}{⇌}S{r}^{2+}+2O{H}^{-}\end{array}$

[Sr2+] = 0.1581 M

[OH] = 2 x 0.1581 M = 0.3162 M

pOH = – log (0.3162) = 0.5

pH = 14 – 0.5 = 13.5

Q.72 What will be the pressure of the gaseous mixture when 0.5 L of H2 at 0.8 bar and 2.0 L of dioxygen at 0.7 bar are introduced in a 1L vessel at 27°C?

Ans.

Let the partial pressure of H2 in the vessel be PH2

Given:

Alternatively,

$\begin{array}{l}Partial\text{pressure of Hydrogen in 1 L container is calculated according to Boyle’s law}\\ {P}_{1}{V}_{1}={P}_{2}{V}_{2}⇒0.8×0.5={P}_{2}×1\\ ⇒\text{}\therefore \text{}{P}_{2}={p}_{{H}_{2}}=0.4\text{bar}\\ \text{Similarly}\dots \text{Partial pressure of dioxygen in 1 L container}\\ {P}_{1}{V}_{1}={P}_{2}{V}_{2}⇒0.7×2.0={P}_{2}×1\\ ⇒\text{}\therefore \text{}{P}_{2}={p}_{{O}_{2}}=1.4\text{bar}\\ \text{Therefore}\dots {\text{P}}_{total}={p}_{{H}_{2}}+{p}_{{O}_{2}}=0.4+1.4=1.8\text{bar}\end{array}$

Total pressure of the gaseous mixture in the vessel is 1.8 bar

Q.73 Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at –10.0°C.

fusH = 6.03 kJ mol–1 at 0°C.

Cp[H 2O(l)] = 75.3 J mol–1 K–1

Cp[H 2O(s)] = 36.8 J mol–1 K–1

Ans.

Total enthalpy change involved in the transformation is the sum of the following changes:

(a) Energy change involved in the transformation of 1 mol of water at 10°C to 1 mol of water at 0°C.

(b) Energy change involved in the transformation of 1 mol of water at 0° to 1 mol of ice at 0°C.

(c) Energy change involved in the transformation of 1 mol of ice at 0°C to 1 mol of ice at –10°C.

Total ∆H=Cp[H2OCl]∆T+∆H freezing +Cp [H2O(S)]∆T

= (75.3 J mol–1 K–1) (0 – 10)K + (–6.03 × 103 J mol–1) + (36.8 J mol–1 K–1) (–10 – 0)K

= –753 J mol–1 – 6030 J mol–1 – 368 J mol–1

= –7151 J mol–1

= –7.151 kJ mol–1

Hence, the enthalpy change involved in the transformation is –7.151 kJ mol–1.

Q.74 The ionization constant of propanoic acid is 1.32 ×10–5. Calculate the degree of ionization of the acid in its 0.05 M solution and also its pH. What will be its degree of ionization if the solution is 0.01 M in HCl also?

Ans.

Let the degree of ionization of propanoic acid be α.

Assume propionic acid as HA, we have:

$\begin{array}{l}\underset{\begin{array}{l}\\ 0.05-0.05\alpha \end{array}}{HA+{H}_{2}O}\underset{}{⇌}\underset{\begin{array}{l}\\ 0.05\alpha \end{array}}{{H}_{3}{O}^{+}}+\underset{\begin{array}{l}\\ 0.05\alpha \end{array}}{{A}^{-}}\\ \text{Let us assumeas 0}\text{.05}-0.05\alpha \text{as}0.0\text{5}.\\ {K}_{a}=\frac{\left[{H}_{3}{O}^{+}\right]\left[{A}^{-}\right]}{\left[HA\right]}\\ =\frac{\left(0.05\alpha \right)\left(0.05\alpha \right)}{0.05}=0.05{\alpha }^{2}\\ \alpha =\sqrt{\frac{{K}_{a}}{0.05}}\\ =\sqrt{\frac{1.32×{10}^{-5}}{.05}}\\ =1.63×{10}^{-2}\end{array}$

[H3O+] = 0.05 α = 0.05 x 1.63 x 10–2 = 8.15 x 10–4 M

pH = – log [H+]

= – log [H3O+]

= – log (8.15 x 10–4)

= 3.09

In the presence of 0.1 M of HCl, let α’ be the degree of ionization.

[H3O+] = 0.01

[A] = 0.05 α’

[HA]= 0.05

$\begin{array}{l}{K}_{a}=\frac{\left(0.01\right)\left(0.05\alpha ‘\right)}{0.05}\\ ⇒1.32×{10}^{-5}=0.01×\alpha ‘\\ ⇒\alpha ‘=1.32×{10}^{-3}\end{array}$

Q.75 The pH of 0.1 M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.

Ans.

Concentration c = 0.1M

pH = 2.34

– log [H+] = 2.34

or [H+] = antilog (-2.34)

or [H+] = 4.5 x 10–3

Also [H+] = c . α

⇒ 4.5 x 10–3 = 0.1 x α

⇒ α = 45 x 10–3= 0.045

Ka = c . α2

= 0.1 (45 x 10–3)2

= 2.02 x 10–4

Q.76 Enthalpy of combustion of carbon to CO2 is –393.5 kJ mol–1. Calculate the heat released upon formation of 35.2 g of CO2 from carbon and dioxygen gas.

Ans.

Formation of CO2 from carbon and dioxygen gas can be represented as:

C(s )+O2(g) →CO2(g) H=-393.5 kJ mol-1

(1 mole = 44 g)

Heat released on formation of 44 g CO2 = 393.5 kJ mol–1

Heat released on formation of 35.2 g CO2

= 314.8 kJ mol–1

Q.77 The ionization constant of nitrous acid is 4.5 × 10–4. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.

Ans.

NaNO2 is the salt of a strong base (NaOH) and a weak acid (HNO2).

$\underset{\begin{array}{l}\\ 0.04-x\end{array}}{N{O}_{2}^{-}+{H}_{2}O}\underset{}{⇌}\underset{\begin{array}{l}\\ x\end{array}}{HN{O}_{2}}+\underset{\begin{array}{l}\\ x\end{array}}{O{H}^{-}}$ $\begin{array}{l}{K}_{h}=\frac{\left[HN{O}_{2}\right]\left[O{H}^{-}\right]}{\left[N{O}_{2}^{-}\right]}\\ ⇒\frac{{K}_{w}}{{K}_{a}}=\frac{{10}^{-14}}{4.5×{10}^{-4}}=0.22×{10}^{-10}\end{array}$

Suppose x mole of salt undergoes hydrolysis, the concentration of various species present in the solution will be:

[NO2]=0.04 – x ~ 0.04

[HNO2] = x

[OH] = x

$\begin{array}{l}{K}_{h}=\frac{\left[HN{O}_{2}\right]\left[O{H}^{-}\right]}{\left[N{O}_{2}^{-}\right]}\\ {K}_{h}=\frac{{x}^{2}}{0.04}=0.22×{10}^{-10}\\ ⇒{x}^{2}=0.0088×{10}^{-10}\\ ⇒x=0.093×{10}^{-5}\end{array}$

[OH] = 0.093 x 10–5 M

[H+]= Kw /(0.093 x 10–5)

= 10–14 / (0.093 x 10–5)

= 10.75 x 10–9 M

pH = – log (10.75 x 10–9) = 7.96

Therefore, degree of hydrolysis h

$\begin{array}{l}h=\frac{x}{0.04}=\frac{0.093×{10}^{-5}}{0.04}\\ =2.325×{10}^{-5}\end{array}$

Q.78 A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.

Ans.

pH = – log [H+]

⇒3.44 = – log [H+]

⇒[H+] = – antilog pH

⇒[H+] = – antilog (3.44)

⇒[H+] = 3.63 x 10–4

$\begin{array}{l}{K}_{h}=\frac{\left[pyridinium\text{chloride}\right]\left[{H}^{+}\right]}{\left[pyridinium\text{hydrochloride}\right]}\\ ⇒{K}_{h}\frac{{\left(3.63×{10}^{-4}\right)}^{2}}{0.02}=6.6×{10}^{-6}\\ ⇒{K}_{h}=\frac{{K}_{w}}{{K}_{a}}=\frac{{10}^{-14}}{6.6×{10}^{-6}}=1.51×{10}^{-9}\end{array}$

Q.79 Predict if the solutions of the following salts are neutral, acidic or basic: NaCl, KBr, NaCN, NH4NO3, NaNO2 and KF.

Ans.

NaCN, NaNO2, KF solutions are basic, as they are salts of strong base and weak acid.

NaCl, KBr solutions are neutral, as they are salts of strong base and strong acid.

NH4NO3 solution is acidic, as it is a salt of strong acid and weak base.

Q.80 The ionization constant of chloroacetic acid is 1.35 × 10–3. What will be the pH of 0.1 M acid and its 0.1 M sodium salt solution?

Ans.

It is given that Ka for ClCH2COOH is 1.35 × 10–3

$\begin{array}{l}{K}_{a}=c{\alpha }^{2}\\ \alpha =\sqrt{\frac{{K}_{a}}{C}}\\ \text{Let us put}\text{the values in the above expression}\\ \alpha =\sqrt{\frac{1.35×{10}^{-3}}{0.1}}\\ ⇒\alpha =\sqrt{1.35×{10}^{-2}}\end{array}$

= 0.116

[H+] = c . a

= 0.1 x 0.116

= 0.0116

$⇒pH=-\mathrm{log}\left[{H}^{+}\right]$

= – log (.0116)

=1.94

ClCH2COONa is the salt of a weak acid i.e., ClCH2COOH and a strong base i.e., NaOH.

$\begin{array}{l}pH=-\frac{1}{2}\left[\mathrm{log}{K}_{w}+\mathrm{log}{K}_{a}-\mathrm{log}c\right]\\ ⇒ph=-\frac{1}{2}\left[\mathrm{log}{10}^{-14}+\mathrm{log}\left(1.35×{10}^{-3}\right)-\mathrm{log}0.1\right]\\ ⇒pH=-\frac{1}{2}\left[-14+\left(-3+0.1303\right)-\left(-1\right)\right]\\ =7.94\end{array}$

Q.81 Ionic product of water at 310 K is 2.7 × 10–14. What is the pH of neutral water at this temperature?

Ans.

Ionic product of water

$\begin{array}{l}{K}_{w}=\left[{H}^{+}\right]\left[O{H}^{-}\right]\\ Since,\\ \left[{H}^{+}\right]=\left[O{H}^{-}\right]\end{array}$

Thus, Kw = [H+]2 and Kw at 310 K is 2.7 × 10–14

⇒2.7 × 10–14 = [H+]2

⇒ [H+] = 1.64 x 10–7

pH = – log [H+]

= – log (1.64 x 10–7)

= 6.78

Hence, the pH of neutral water is 6.78.

Q.82 Density of a gas is found to be 5.46 g/dm at 27 °C at 2 bar pressure. What will be its density at STP?

Ans.

Given,

$\begin{array}{l}{d}_{1}=5.46\text{g}/d{m}^{3}\\ {p}_{1}=2\text{bar}\\ {\text{T}}_{1}=27ºC=\left(27+273\right)K=300\text{K}\\ {p}_{2}=1\text{bar}\\ {\text{T}}_{2}=273\text{K}\\ {\text{d}}_{2}=?\end{array}$

The density (d2) of the gas at STP can be calculated using the equation,

$\begin{array}{l}d=\frac{Mp}{RT}\\ \therefore \text{}\frac{{d}_{1}}{{d}_{2}}=\frac{\frac{M{p}_{1}}{R{T}_{1}}}{\frac{M{p}_{2}}{R{T}_{2}}}\\ ⇒\text{}\frac{{d}_{1}}{{d}_{2}}=\frac{{p}_{1}{T}_{2}}{{p}_{2}{T}_{1}}\\ ⇒\text{}{d}_{2}=\frac{{p}_{2}{T}_{1}{d}_{1}}{{p}_{1}{T}_{2}}\\ =\frac{1×300×5.46}{2×273}\\ =3\text{g}d{m}^{-3}\end{array}$

Density of the gas at STP will be 3 g dm–3

Q.83 Calculate the pH of the resultant mixtures:

a) 10 mL of 0.2 M Ca(OH)2+ 25 mL of 0.1 M HCl

b) 10 mL of 0.01 M H2SO4+ 10 mL of 0.01 M Ca(OH)2

c) 10 mL of 0.1 M H2SO4+ 10 mL of 0.1 M KOH

Ans.

(a)

10 mL of 0.2 M Ca(OH)2 = 10 x 0.2 = 2 millimoles of Ca(OH)2

25 mL of 0.1M HCl = 25 x 0.1 millimoles = 2.5 millimoles of HCl

$Ca{\left(OH\right)}_{2}2HCl\stackrel{}{\to }CaC{l}_{2}+2{H}_{2}O$

2 millimoles of HCl reacts with = 1 millimoles of Ca(OH)2

1 millimoles of HCl reacts with = 1/2 millimoles of Ca(OH)2

2.5 millimoles of HCl reacts with = ½ x 2.5 = 1.25 millimoles of Ca(OH)2

Amount of Ca(OH)2 left = 2 – 1.25 = 0.75 millimoles

Total volume of the solution (10 + 25) mL = 35 mL

Molarity of Ca(OH)2 in the solution = (0.75/35)M = 0.0214 M

[OH] = 2 x 0.0214 M = 0.0428 M = 4.28 x 10–2 M

pOH = – log (4.28 x 10–2) = 2 – 0.6314 » 1.37

pH = 14 – 1.37 = 12.63

(b) 10 mL 0.01 M H2SO4 = 10 x 0.01 millimoles of H2SO4 = 0.1 millimole

10 mL of 0.01 M Ca(OH)2 = 10 x 0.01 millimoles of Ca(OH)2 = 0.1 millimole

$Ca{\left(OH\right)}_{2}+{H}_{2}S{O}_{4}\stackrel{}{\to }CaS{O}_{4}+2{H}_{2}O$

1 mole of Ca(OH)2 reacts with 1 mole of H2SO4.

(c) Thus, 0.1 millimole of Ca(OH)2 reacts completely with 0.1 millimole of H2SO4. The solution will be neutral with pH value as 7.0.10 mL 0.1 M H2SO4 = 1 millimole of H2SO4

10 mL 0.1 M KOH = 1 millimole of KOH

$2KOH+{H}_{2}S{O}_{4}\stackrel{}{\to }{K}_{2}S{O}_{4}+2{H}_{2}O$

2 millimole of KOH reacts with = 1 millimole of H2SO4

1 millimole of KOH reacts with = ½ = 0.5 millimole of H2SO4

Amount of H2SO4 left = (1 – .05) = 0.5 millimole

Volume of the reaction mixture = 10 + 10 = 20 mL

Molarity of H2SO4 in the solution = 0.5/20 = 2.5 x 10–2 M

[H+] = 2 x (2.5 x 102) = 5 x 10–2

pH = – log (5 x 10–2) = 2 – 0.699 = 1.3

Q.84 34.05 mL of phosphorus vapour weighs 0.0625 g at 546 °C and 0.1 bar pressure. What is the molar mass of phosphorus?

Ans.

p = 0.1 bar

V = 34.05 mL = 34.05 × 10–3 L = 34.05 × 10–3 dm3

R= 0.083 bar dm3 K–1 mol–1

T = 546°C = (546 + 273) K = 819 K

According to the ideal gas equation:

$\begin{array}{l}pV=nRT\\ ⇒n=\frac{pV}{RT}\\ =\frac{0.1×34.05×{10}^{-3}}{0.083×819}\\ =5.01×{10}^{-5}\text{mol}\\ \text{Molar mass of phosphorus}=\frac{0.0625}{5.01×{10}^{-5}}=124.75{\text{g mol}}^{-1}\end{array}$

The molar mass of phosphorus is 124.75 g mol–1

Q.85 A student forgot to add the reaction mixture to the round bottomed flask at 27 °C but instead he/she placed the flask on the flame. After a lapse of time, he realized his mistake, and using a pyrometer he found the temperature of the flask was 477 °C. What fraction of air would have been expelled out?

Ans.

If the volume of the round bottomed flask be V.

Then, the volume of air inside the flask at 27° C is V.

Let volume of air inside flask will be equal to volume of flask and let that volume be V

T = 27°C = 300 K

V2 =?

T2 = 477° C = 750 K

According to Charles’s law,

$\begin{array}{l}\frac{{V}_{1}}{{T}_{1}}=\frac{{V}_{2}}{{T}_{2}}\\ ⇒{V}_{2}=\frac{{V}_{1}{T}_{2}}{{T}_{1}}\\ =\frac{750\text{V}}{300}\\ =2.5\text{V}\end{array}$

The volume of air expelled out = 2.5 V – V = 1.5 V

$Fraction\text{of air expelled out}=\frac{1.5V}{2.5V}=\frac{3}{5}$

Q.86 Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298 K from their solubility product constants given: Ag2CrO4 = 1.1 x 10–12,

BaCrO4 = 1.2 x 10–10, Fe(OH)3 = 1.0 x 10–38,

PbCl2 = 1.6 x 10–5, Hg2I2 = 4.5 x 10–29

Determine also the molarities of individual ions.

Ans.

1. Silver chromate:

$\begin{array}{l}A{g}_{2}Cr{O}_{4}\underset{}{⇌}2A{g}^{+}+Cr{O}_{4}^{2-}\\ {K}_{sp}={\left[A{g}^{+}\right]}^{2}\left[Cr{O}_{4}^{2-}\right]\end{array}$

Let the solubility of Ag2CrO4 be s.

For [Ag+] = 2s

[CrO42–] = s

Then,

Ksp = (2s) 2.s = 4 s3

⇒ 1.1 x 10–12 = 4 s3

⇒ s = 0.65 x 10–4 M

Molarity of Ag+ = 2 s

= 2 x (0.65 x 10–4)

= 1.3 x 10–4 M

Molarity of CrO42– = s = 0.65 x 10–4 M

1. Barium chromate:

$\begin{array}{l}BaCr{O}_{4}\underset{}{⇌}B{a}^{2+}+Cr{O}_{4}^{2-}\\ {K}_{sp}=\left[B{a}^{+}\right]\left[Cr{O}_{4}^{2-}\right]\end{array}$ ${K}_{sp}=\left[B{a}^{+}\right]\left[Cr{O}_{4}^{2-}\right]$

The solubility of BaCrO4 is s.

Thus, [Ba2+] = s

and [CrO42–] = s

Ksp = s . s = s2

⇒ s2 = 1.2 x 10–10

⇒ s = 1.09 x 10–5 M

Molarity of Ba2+ = Molarity of CrO42– = 1.09 x 10–5 M

1. Ferric hydroxide:

$\begin{array}{l}Fe{\left(OH\right)}_{3}\underset{}{⇌}F{e}^{3+}+3O{H}^{-}\\ {K}_{sp}=\left[F{e}^{3+}\right]{\left[O{H}^{-}\right]}^{3}\end{array}$

Let the solubility product of Fe(OH)3 be s.

Thus, [Fe3+] = s

and [OH] = 3s

∴ Ksp = s. (3s)3 = 27 s4

⇒ Ksp = 27 s4

⇒ 1.0 x 10–38 = 27 s4

⇒ 0.037 x 10–38 = s4

⇒ s = 1.39 x 10–10 M

Molarity of Fe3+ = s = 1.39 x 10–10 M

Molarity of OH= 3s = 3 x 1.39 x 10–10 M

= 4.17 x 10–10 M

$\begin{array}{l}PbC{l}_{2}\underset{}{⇌}P{b}^{2+}+2C{l}^{-}\\ {K}_{sp}=\left[P{b}^{2+}\right]{\left[C{l}^{-}\right]}^{2}\end{array}$

Let the solubility product of PbCl2 be s.

Thus, [Pb2+] = s

and [Cl] = 2s

∴ Ksp = s. (2s)2

⇒ Ksp = 4 s3

⇒ 1.6 x 10–5 = 4 s3

⇒ s3 = 4 x 10–6

⇒ s = 1.58 x 10–2 M

Molarity of Pb2+ = s = 1.58 x 10–2 M

Molarity of Cl = 2s = 2 x 1.58 x 10–2 M

= 3.16 x 10–2 M

5. Mercurous iodide:

$\begin{array}{l}H{g}_{2}{I}_{2}\underset{}{⇌}H{g}_{2}^{2+}+2{I}^{-}\\ {K}_{sp}=\left[H{g}_{2}^{2+}\right]{\left[{I}^{-}\right]}^{2}\end{array}$

Let the solubility product of Hg2I2 be s.

[Hg22+] = s

and [I] = 2s

∴ Ksp= s . (2s)2

⇒ Ksp= 4 s3

⇒4.5 x 10–29 = 4 s3

⇒ s = 2.24 x 10–10 M

Molarity of [Hg22+] = s = 2.24 x 10–10 M

Molarity of [I] = 2s = 2 x 2.24 x 10–10 M

= 4.48 x 10–10 M

Q.87 Enthalpies of formation of CO(g), CO2(g), N 2 O(g) and N 2 O4(g) are –110 kJ mol–1, – 393 kJ

mol–1, 81 kJ mol–1 and 9.7 kJ mol–1 respectively. Find the value of ∆rH for the reaction:

N2O4(g) + 3CO(g) → N2O(g) + 3CO2(g)

Ans.

rH for a reaction is defined as the difference between ∆fH value of products and ∆fH

value of reactants.

Δ rH = ∑ ΔH (Products) – ∑ ΔH (reactants)

For the given reaction,

N2O4(g) + 3CO(g) N2O(g) + 3CO2(g)

ΔrH =[{ ΔH(N2O)+3 ΔH (CO2)}-{ ΔH(N2O4)+3 ΔH(CO)}]

Substituting the values of ∆fH for N2O, CO2, N2O4, and CO from the question, we get:

ΔrH =[{81 kJ mol-1 +3(-393)kJ mol-1}-{9.7 kJ mol-1 +3(-110)kJ mol-1}]

ΔrH =-777.7 kJ mol-1

Hence, the value of ∆rH for the reaction is -777.7 kJ mol-1

Q.88 The solubility product constant of Ag2CrO4 and AgBr are 1.1 × 10–12 and 5.0 × 10–13 respectively. Calculate the ratio of the molarities of their saturated solutions.

Ans.

Let the solubility of Ag2CrO4 = s

$\begin{array}{l}A{g}_{2}Cr{O}_{4}\underset{}{⇌}2A{g}^{+}+Cr{O}_{4}^{2-}\\ {K}_{sp}={\left[A{g}^{+}\right]}^{2}\left[Cr{O}_{4}^{2-}\right]\end{array}$

Let the solubility of Ag2CrO4 be s.

For [Ag+] = 2s

[CrO42–] = s

Then,

Ksp= (2s2). s = 4 s3

⇒ 1.1 x 10–12 = 4 s3

⇒ s = 0.65 x 10–4 M = 6.5 x 10–5 M

Let s’ be the solubility of AgBr.

$AgBr\left(s\right)\underset{}{⇌}A{g}^{+}+B{r}^{-}$

Thus, [Ag+] = [Br] = s

⇒Ksp = (s’)2 = 5.0 x 10–13

⇒ s’ = 7.07 x 10–7 M

Therefore, the ratio of the molarities of their saturated solution is

$\frac{s}{s‘}=\frac{6.5×{10}^{-5}M}{7.07×{10}^{-7}M}=91.9$

Q.89 Calculate the temperature of 4.0 mol of a gas occupying 5 dm3at 3.32 bar.

(R= 0.083 bar dm3 K–1 mol–1)

Ans.

Given,

n = 4.0 mol

V = 5 dm3

p = 3.32 bar

R= 0.083 bar dm3 K–1 mol–1

Using the ideal gas equation:

pV = nRT

$\begin{array}{l}⇒T=\frac{pV}{nR}\\ =\frac{3.32×5}{4×0.083}\\ =50\text{K}\end{array}$

Temperature is 50 K.

Q.90 Calculate the total number of electrons present in 1.4 g of dinitrogen gas.

Ans.

Molar mass of dinitrogen (N2) = 28 g mol–1

$\begin{array}{l}1.4{\text{g of N}}_{2}=\frac{1.4}{28}=0.05\text{mol}\\ =\text{0}\text{.05×6}{\text{.02×10}}^{23}\text{number of molecules}\\ \text{3}{\text{.01×10}}^{22}\text{number of molecules}\end{array}$

1 molecule of N2 contains 14 electrons.
3.01 × 1022 molecules of N2 contains = 14 × 3.01 × 1022
= 4.214 × 1023 electrons

Q.91 Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate Ksp= 7.4 × 10–8).

Ans.

When we mix equal volumes of sodium iodate and cupric chlorate solutions, then the molar concentrations of both solutions are reduced to half i.e., 0.001 M.

Considering the ionization of both compounds:

$\begin{array}{l}\underset{\begin{array}{l}\\ 0.001\text{M}\end{array}}{NaI{O}_{3}}\stackrel{}{\to }N{a}^{+}+\underset{\begin{array}{l}\\ 0.001\text{M}\end{array}}{I{O}_{3}^{-}}\\ \underset{\begin{array}{l}\\ 0.001\text{M}\end{array}}{Cu{\left(Cl{O}_{3}\right)}_{2}}\stackrel{}{\to }C{u}^{2+}+\underset{\begin{array}{l}\\ 0.001\text{M}\end{array}}{2C{l}_{3}^{-}}\\ \text{The solubility equilibrium for copper iodate can be written as}:\\ Cu{\left(I{O}_{3}\right)}_{2}\left(aq\right)\underset{}{⇌}C{u}^{2+}\left(aq\right)+2I{O}_{3}^{-}\left(aq\right)\end{array}$

Ionic product of copper iodate:

= [Cu2+] [IO3]2

= (0.001) (0.001)2

= 1 x 10–9

The Ksp value for copper iodate, Ksp = 7.4 × 10–8

Since the ionic product (1 × 10–9) is less than Ksp (7.4 × 10–8), precipitation will not occur.

Q.92 Given

N2(g)+3H2(g) → 2NH3(g) ∆rHθ = –92.4 kJ mol–1

What is the standard enthalpy of formation of NH3 gas?

Ans.

Standard enthalpy of formation of a compound is the change in enthalpy that takes place during the formation of 1 mole of a substance in its standard form from its constituent elements in their standard state.Re-writing the given equation for 1 mole of NH3(g),

$\frac{1}{2}{N}_{2\left(g\right)}+\frac{3}{2}{H}_{2\left(g\right)}\to N{H}_{3\left(g\right)}$

...Standard enthalpy of formation of NH3(g)

= ½ ∆rHθ

= ½ (–92.4 kJ mol–1)

= –46.2 kJ mol–1

Q.93 How much time would it take to distribute one Avogadro number of wheat grains, if 1010grains are distributed each second?

Ans.

Avogadro number = 6.02 × 1023

10 grains are distributed in = 1sec,

Therefore, 6.02 × 1023 will be calculated as:Time required

$\begin{array}{l}\frac{6.02×{10}^{23}}{{10}^{10}}S\\ =6.02×{10}^{23}S\\ =\frac{6.02×{10}^{23}}{60×60×24×365}years\\ =1.909×{10}^{6}years\\ Time\text{taken}=1.909×{10}^{6}years\end{array}$

Q.94 The ionization constant of benzoic acid is 6.46 × 10–5 and Ksp for silver benzoate is 2.5 × 10–13. How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water?

Ans.

Given,

pH = – log [H+] = 3.19

⇒[H+] = antilog (–3.19)

= 6.457 x 10–4 M

$\begin{array}{l}{C}_{6}{H}_{5}COOH\left(aq\right)\underset{}{⇌}{C}_{6}{H}_{5}CO{O}^{-}\left(aq\right)+{H}^{+}\left(aq\right)\\ {K}_{a}=\frac{\left[{C}_{6}{H}_{5}CO{O}^{-}\left(aq\right)\right]\left[{H}^{+}\left(aq\right)\right]}{\left[{C}_{6}{H}_{5}COOH\left(aq\right)\right]}\\ ⇒\frac{\left[{C}_{6}{H}_{5}COOH\left(aq\right)\right]}{\left[{C}_{6}{H}_{5}CO{O}^{-}\left(aq\right)\right]}=\frac{\left[{H}^{+}\left(aq\right)\right]}{{K}_{a}}=\frac{6.46×{10}^{-4}}{6.46×{10}^{-5}}=10\end{array}$

Let the solubility of C6H5COOAg be x mol/L

Then,

[Ag+] = x

⇒ [C6H5COOH] + [C6H5COO] = x

⇒ 10 [C6H5COO] + [C6H5COO] = x

⇒11 [C6H5COO] = x

⇒ [C6H5COO] = x/11

$\begin{array}{l}{K}_{sp}=\left[A{g}^{+}\right]\left[{C}_{6}{H}_{5}CO{O}^{-}\right]\\ ⇒2.5×{10}^{-13}=x.\left(x/11\right)\end{array}$

⇒ x = 1.66 x 10–6 mol/L

Thus, the solubility of silver benzoate in a pH 3.19 solution is 1.66 × 10–6 mol/L.

Let the solubility of C6H5COOAg be y mol/L in pure water.

[Ag+] =y M

and [CH3COO]= y M

$\begin{array}{l}{K}_{sp}=\left[A{g}^{+}\right]\left[{C}_{6}{H}_{5}CO{O}^{-}\right]\\ ⇒y=\sqrt{Ksp}=\sqrt{\left(2.5×{10}^{-13}\right)}=5×{10}^{-7}mol/L\\ \frac{x}{y}=\frac{1.66×{10}^{-6}}{5×{10}^{-7}}=3.32\end{array}$

Hence, C6H5COOAg is approximately 3.32 times more soluble in a low pH solution.

Q.95 Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 dm3 at 27°C. R = 0.083 bar dm3 K–1 mol–1

Ans.

Mass of dioxygen (O2) = 8 g (Given)

$\begin{array}{l}No.{\text{of moles of O}}_{2}=\frac{Mass}{M.\text{}wt}\\ {O}_{2}=\frac{8}{32}=0.25\text{mole}\\ \text{Mass of dihydrogen}\left({\text{H}}_{\text{2}}\right)\text{}=\text{4 g (Given)}\\ No.{\text{of moles of H}}_{2}=\frac{Mass}{M.\text{}wt}\\ Number{\text{of moles of H}}_{2}=\frac{4}{2}=2\text{mole}\end{array}$

Total number of moles in the mixture = 0.25 + 2 = 2.25 mole

V = 1 dm3

n = 2.25 mol

R= 0.083 bar dm3 K–1 mol–1

T = 27°C = 300 K

Total pressure (p) is:

pV = nRT

$\begin{array}{l}Or,\text{}p=\frac{nRT}{V}\\ =2.25×0.083×3001\\ =56.025\text{bar}\end{array}$

The total pressure of the mixture is 56.025 bar.

Q.96 Pay load is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the pay load when a balloon of radius 10 m, mass 100 kg is filled with helium at 1.66 bar at 27°C. (Density of air = 1.2 kg m–3 and R = 0.083 bar dm3 K–1Mol–1).

Ans.

Radius of the balloon, r = 10 m

Volume of a sphere is given as:

$\begin{array}{l}V=\frac{4}{3}\pi {r}^{3}\\ \therefore \text{}\text{Volume of ballon}=\frac{4}{3}\pi {r}^{3}\\ =\frac{4}{3}×\frac{22}{7}×{10}^{3}\\ =4190.5{\text{m}}^{3}\left(approx\right)\end{array}$

∴ The volume of the displaced air is 4190.5 m3.

Density of air = 1.2 kg m–3

Mass = Volume × density
Mass of displaced air = 4190.5 × 1.2 kg = 5028.6 kg

Mass of helium (m) inside the balloon is given by,

$\begin{array}{l}m=\frac{MpV}{RT}\\ Here,\\ M=4×{10}^{-3}kg{\text{mol}}^{-1}\\ p=1.66\text{bar}\end{array}$

V=Volume of the balloon
=4190.5 m3

$\begin{array}{l}R=0.083{\text{bar dm}}^{3}{K}^{-1}mo{l}^{-1}\\ T=27ºC=300\text{}K\\ m=\frac{4×{10}^{-3}×1.66×4190.5×{10}^{3}}{0.083×300}\end{array}$

=1117.5 kg (approx)

Now, total mass of the balloon filled with helium = (100 + 1117.5) kg

= 1217.5 kg

Hence, pay load = (5028.6 – 1217.5) kg

= 3811.1 kg

Hence, the pay load of the balloon is 3811.1 kg.

Q.97 Calculate the standard enthalpy of formation of CH3OH(l) from the following data:

$\begin{array}{l}C{H}_{3}O{H}_{\left(I\right)}+\frac{2}{3}{O}_{2\left(g\right)}\to C{O}_{2}{O}_{\left(I\right)};\text{}{?}_{r}{H}^{\theta }=-726{\text{kJ mol}}^{-1}\\ {C}_{\left(g\right)}+{O}_{2\left(g\right)}\to C{O}_{2\left(g\right)};\text{}{}_{}\end{array}$

Δ

c H θ =393 kJ mol 1 H 2( g ) + 1 2 O 2( g ) H 2 O ( I ) ;

Δ

Ans.

The reaction that takes place during the formation of CH3OH(l) can be written as:

$C\left(s\right)+2{H}_{2}O\left(g\right)+\frac{1}{2}{O}_{2\left(g\right)}\to C{H}_{3}O{H}_{\left(I\right)}\text{}\left(1\right)$

The reaction (1) can be obtained from the given reactions by following the algebraic

calculations as:

Equation (ii) + 2 × equation (iii) – equation (i)

fHθ [CH3OH(l)] = ∆cHθ + 2∆fHθ [H2O(l)] – ∆rHθ

= (–393 kJ mol–1) + 2(–286 kJ mol–1) – (–726 kJ mol–1)

= (–393 – 572 + 726) kJ mol–1

fHθ [CH3OH(l)] = –239 kJ mol–1

Q.98 Calculate the volume occupied by 8.8 g of CO2 at 31.1°C and 1 bar pressure.

R= 0.083 bar L K–1 mol–1.

Ans.

Since,

$\begin{array}{l}pV=\frac{m}{M}RT\\ ⇒V=\frac{mRT}{Mp}\end{array}$

Here,
m = 8.8 g
R = 0.083 bar LK–1 mol–1
T = 31.1°C = 304.1 K
M = 44 g
p = 1 bar

$Volume\left(V\right)=\frac{8.8×0.083×304.1}{44×1}$

=5.04806L
=5.05L
The volume occupied is 5.05 L.

Q.99 What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide? (For iron sulphide, Ksp= 6.3 × 10–18).

Ans.

Let the maximum concentration of each solution be x mol/L. After mixing, the concentrations of each solution will be reduced to half i.e., (x/2) mol/L.

$\begin{array}{l}\therefore \left[FeS{O}_{4}\right]=\left[N{a}_{2}S\right]=\frac{x}{2}M\\ \therefore \left[F{e}^{2+}\right]=\therefore \left[FeS{O}_{4}\right]=\frac{x}{2}M\\ \therefore \left[{S}^{2-}\right]=\left[N{a}_{2}S\right]=\frac{x}{2}M\\ FeS\left(s\right)\underset{}{⇌}F{e}^{2+}\left(aq\right)+{S}^{2-}\left(aq\right)\\ {K}_{sp}=\left[F{e}^{2+}\right]\left[{S}^{2-}\right]\\ ⇒6.3×{10}^{-18}=\frac{x}{2}×\frac{x}{2}\\ ⇒6.3×{10}^{-18}=\frac{{x}^{2}}{4}\\ ⇒x=5.02×{10}^{-9}\end{array}$

If the concentrations of both solutions are equal to or less than 5.02 × 10–9 M, then there will be no precipitation of iron Sulphide.

Q.100 Calculate the enthalpy change for the process

CCl4(g) → C(g) + 4Cl(g)

and calculate bond enthalpy of C–Cl in CCl4(g).

vapHθ (CCl4) = 30.5 kJ mol–1.

fHθ (CCl4) = –135.5 kJ mol–1.

aHθ (C) = 715.0 kJ mol–1, where ∆aHθ is enthalpy of atomisation

aHθ (Cl2) = 242 kJ mol–1

Ans.

The chemical equations implying to the given values of enthalpies are:

CCI4(l) → CCI4(g)vapHθ = 30.5 kJ mol–1

C(s) → C(g)aHθ = 715.0 kJ mol–1

Cl2(g) → 2Cl(g)aHθ = 242 kJ mol–1

C(g)+Cl(g) → CCI4(g)fH = –135.5 kJ mol–1

Enthalpy change for the given processCCI4(g) → C(g)+4CI(g)’ can be calculated using the following algebraic calculations as:

Equation (ii) + 2 × Equation (iii) – Equation (i) – Equation (iv)

∆H = ∆aHθ(C) + 2∆aHθ (Cl2) – ∆vapHθ – ∆fH

= (715.0 kJ mol–1) + 2(242 kJ mol–1) – (30.5 kJ mol–1) – (–135.5 kJ mol–1)

∆H = 1304 kJ mol–1

Bond enthalpy of C–Cl bond in CCl4(g)

=326 kJ mol-1

Q.101 2.9 g of a gas at 95 °C occupied the same volume as 0.184 g of dihydrogen at 17 °C, at the same pressure. What is the molar mass of the gas?

Ans.

Volume (V) occupied by dihydrogen is:

$\begin{array}{l}V=\frac{m}{M}\frac{RT}{p}\\ =\frac{0.184}{2}×\frac{R×290}{p}\end{array}$

Let, the molar mass of the unknown gas be M then the volume (V) occupied by it, can be given as:

$\begin{array}{l}V=\frac{m}{M}\frac{RT}{p}\\ =\frac{2.9}{M}\frac{R×368}{p}\\ \text{According to the question,}\\ \frac{0.184}{2}×\frac{R×290}{p}=\frac{2.9}{M}\frac{R×368}{M}\\ ⇒\frac{0.184×290}{2}=\frac{2.9×368}{M}\\ ⇒M=\frac{2.9×368×2}{0.184×290}\\ =40\text{}g{\text{mol}}^{-1}\end{array}$

The molar mass of the gas is 40 g mol–1

Q.102 For an isolated system, ∆U = 0, what will be ∆S?

Ans.

∆S will be positive i.e., greater than zero

Since ∆U = 0, ∆S will be positive and the reaction will be spontaneous.

Q.103 For the reaction at 298 K,

2A + B → C

∆H = 400 kJ mol–1 and ∆S = 0.2 kJ K–1 mol–1

At what temperature will the reaction become spontaneous considering ∆H and ∆S to be

constant over the temperature range?

Ans.

From the expression,

∆G = ∆H – T∆S

Assuming, the reaction at equilibrium, ∆T for the reaction would be:

$\begin{array}{l}T=\frac{\Delta H}{\Delta S}\left(\Delta G=0\text{at equilibrium}\right)\\ \frac{400{\text{kJ mol}}^{-1}}{0.2{\text{kJ K}}^{-1}mo{l}^{-1}}\end{array}$

T = 2000 K

For the reaction to be spontaneous, ∆G must be negative. Hence, for the given reaction

to be spontaneous, T should be greater than 2000 K.

Q.104 What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298 K? (For calcium sulphate, Ksp is 9.1 × 10–6).

Ans.

$\begin{array}{l}CaS{O}_{4}\left(s\right)\underset{}{⇌}C{a}^{2+}\left(aq\right)+S{O}_{4}^{2-}\left(aq\right)\\ {K}_{sp}=\left[C{a}^{2+}\right]\left[S{O}_{4}^{2-}\right]\end{array}$

Let the solubility of CaSO4 be s.

Then, Ksp = s2

⇒ 9.1 x 10–6 = s2

⇒ s = 3.02 x 10–3 mol/L

Molecular mass of CaSO4 = 136 g/mol

Solubility of CaSO4 in gram/L = 3.02 x 10–3 x 136 g/L

= 0.411 g/L

Thus, to dissolve 0.411 g of CaSO4, we need water = 1 L

And to dissolve 1 g of CaSO4, we need water = (1/0.411) L

= 2.43 L

Q.105 A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen.

Ans.

If the weight of dihydrogen be 20 g and the weight of dioxygen be 80 g.

$\begin{array}{l}\text{No}{\text{. of moles of H}}_{2}=\frac{Mass}{M.\text{wt}}\\ The{\text{number of moles of dihydrogen, n}}_{{H}_{2}}=\frac{20}{2}\\ =\text{10 moles and the number moles}\\ {\text{The number of moles of dioxygen, n}}_{{O}_{2}}=\frac{80}{32}=2.5\text{moles}\end{array}$

Total pressure of the mixture, ptotal= 1 bar
Then, partial pressure of dihydrogen,

$\begin{array}{l}{p}_{{H}_{2}}=\frac{{n}_{{H}_{2}}}{{n}_{{H}_{2}}+{n}_{{O}_{2}}}×{p}_{total}\\ =\frac{10}{10+2.5}×1\\ =0.8\text{bar}\end{array}$

The partial pressure of dihydrogen is 0.8 bar.

Q.106 The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen sulphide is 1.0 × 10–19 M. If 10 mL of this is added to 5 mL of 0.04 M solution of the following: FeSO4, MnCl2, ZnCl2 and CdCl2. in which of these solutions precipitation will take place?

Given, Ksp for FeS = 6.3 x 10–18, MnS = 2.5 x 10–13, ZnS = 1.6 x 10–24 and CdS = 8.0 x 10–27

Ans.

For precipitation to take place, it is required that the calculated ionic product exceeds the Ksp value.

10mL of solution containing S2- ion is mixed with 5 mL of metal salt solution.

Before mixing,

$\begin{array}{l}\left[{S}^{2-}\right]=1.0×{10}^{-19}M\text{}\left(\text{Volume}=\text{1}0\text{mL}\right)\\ \left[{M}^{2+}\right]=0.04M\text{}\left(\text{Volume}=\text{5 mL}\right)\end{array}$

After mixing,

[S2–] = 1.0 x 10–19 x (10/15) = 6.67 x 10–20

[M2+]= [Fe3+] = [Mn2+] = [Zn2+] = [Cd2+]

= 0.04 x (5/15) = 1.33 x 10–2 M

∴ Ionic product of each will be=

[M2+][S2–] = (1.33 x 10–2) (6.67 x 10–20)

= 8.87 x 10–22

This ionic product exceeds the Ksp of ZnS and CdS. Therefore, precipitation will occur in ZnCl2 and CdCl2 solutions.

Q.107 (i) Calculate the number of electrons which will together weigh one gram.

(ii) Calculate the mass and charge of one mole of electrons.

Ans.

(i) Mass of one electron = 9.10939 × 10–31kg

Number of electrons that weigh 9.10939 × 10–31kg = 1

Number of electrons that will weigh 1 g or 1 × 10–3 kg

$=\frac{1}{9.10939×{10}^{-31}kg}×1×{10}^{-3}kg$

= 0.1098 × 1028

= 1.098 x 1027

(ii) Mass of one electron = 9.10939 × 10–31kg

Mass of one mole of electron =

(6.022 × 1023) ×(9.10939 ×10–31kg)

= 5.48 × 10–7kg

Charge on one electron = 1.6022 × 10–19coulomb

Charge on one mole of electron

= (1.6022 × 10–19C) (6.022 × 1023)

= 9.65 × 104C

Q.108 (i) Calculate the total number of electrons present in one mole of methane.

(ii) Find

(a) the total number and

(b) the total mass of neutrons in 7 mg of 14C. (Assume that mass of a neutron =

1.675 × 10-27kg).

(iii) Find

(a) the total number and

(b) the total mass of protons in 34 mg of NH3 at STP. Will the answer change if the temperature and pressure are changed?

Ans.

(i) Number of electrons present in 1 molecule of methane (CH4) = {1(6) + 4(1)} = 10

Number of electrons present in 1 mole i.e., 6.023 ×1023 molecules of methane = 6.022 × 1023× 10

= 6.022 × 1024

(ii) (a) Number of atoms of 14C in 1 mole= 6.023 × 1023

Since, 1 atom of 14C contains (14 – 6) i.e., 8 neutrons, the total number of neutrons in 14 g (or 1 mole) of 14C is (6.023 × 1023) × 8.

Therefore, 14 g of 14C contains (6.023 × 1023× 8) neutrons.

Therefore, number of neutrons in 7 mg of 14C is

$=\frac{6.023×{10}^{23}×8×7\text{}mg}{14000\text{mg}}$

= 2.4092 × 1021

(b) Mass of one neutron = 1.67493 × 10–27kg

Mass of total neutrons in 7 mg of 14C =

= (2.4092 × 1021) (1.67493× 10–27kg)

= 4.0352 × 10–6kg

(iii) (a) 1 mole of N of NH3 = {1(14) + 3(1)} g of NH3

= 17 g of NH3 = 6.022× 1023 molecules of NH3

Total number of protons present in 1 molecule of NH3

= {1(7) + 3(1)} =10

Number of protons in 6.023 × 1023 molecules of NH3

= (6.023 × 1023) (10)

= 6.023 × 1024

=17 g of NH3 contains (6.023 × 1024) protons

Number of protons in 34 mg of NH3

$=\frac{6.022×{10}^{24}×34\text{}mg}{17000\text{mg}}$

= 1.2046 × 1022

(b) Mass of one proton = 1.67493 × 10–27kg

Total mass of protons in 34 mg of NH3

= (1.67493 × 10–27kg) (1.2046 × 1022)

= 2.0176 × 10–5kg

The number of sub-atomic particles (protons, electrons, and neutrons) in an atom is independent of temperature and pressure conditions. Hence, the obtained values will remain unchanged if the temperature and pressure is changed.

Q.109 How many neutrons and protons are there in the following nuclei?

${}_{6}^{13}\mathrm{C,}{\text{}}_{8}^{16}\mathrm{O,}{\text{}}_{12}^{24}M\mathrm{g,}{\text{}}_{26}^{56}F\mathrm{e,}{\text{}}_{38}^{88}Sr$

Ans.

Atomic mass of

${}_{6}^{13}C$

= 13

Atomic number = Number of protons = 6

Number of neutrons = (Atomic mass – Atomic number)

= 13 – 6 = 7

Atomic mass of

${}_{8}^{16}O$

=16

Atomic number = 8 and Number of protons = 8

Number of neutrons = (Atomic mass – Atomic number)

= 16 – 8 = 8

Atomic mass of

${}_{12}^{24}Mg$

=24

Atomic number = Number of protons = 12

Number of neutrons = (Atomic mass – Atomic number)

= 24 – 12 = 12

Atomic mass of

${}_{26}^{56}Fe$

=56

Atomic number = Number of protons = 26

Number of neutrons = (Atomic mass – Atomic number)

= 56 – 26 = 30

Atomic mass of

${}_{38}^{88}Sr$

=88

Atomic number = Number of protons = 38

Number of neutrons = (Atomic mass – Atomic number)

= 88 – 38 = 50

Q.110 Write the complete symbol for the atom with the given atomic number (Z) and Atomic mass (A)

(i) Z = 17, A = 35

(ii) Z = 92, A = 233

(iii) Z = 4, A = 9

Ans.

Q.111 For the reaction,

2Cl(g) → Cl2(g), what are the signs of ∆H and ∆S ?

Ans.

∆H and ∆S are negative

The given reaction represents the formation of chlorine molecule from chlorine atoms.

Here, bond formation is taking place. Therefore, energy is being released. Hence, ∆H is negative.

Also, two moles of atoms have more randomness than one mole of a molecule. Since spontaneity is decreased, ∆S is negative for the given reaction.

Q.112 Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm. Calculate the frequency (ν) and wave number

$\left(\overline{v}\right)$

of the yellow light.

Ans.

λ= c/ν

or, ν = c/ λ ——— (1)

Here,

ν = frequency of yellow light

c = velocity of light in vacuum = 3 × 108 m/s

λ= wavelength of yellow light = 580 nm = 580 × 10–9m

Substituting the values in expression (1):

$v=\frac{3×{10}^{8}}{580×{10}^{-9}}=5.17×{10}^{14}{s}^{-1}$

Thus, frequency of yellow light emitted from the sodium lamp is = 5.17 × 1014s–1.

Wave number of yellow light, ν = 1/λ.

$\overline{v}=\frac{1}{{580 x 10}^{-9}}=1.72×{10}^{6}{m}^{-1}$

Q.113 Find energy of each of the photons which

(i) correspond to light of frequency 3× 1015 Hz

(ii) have wavelength of 0.50 Å.

Ans.

(i) Energy (E) of a photon is given by the expression,

E = hν

Where,

h = Planck’s constant = 6.626 × 10–34Js

ν= frequency of light = 3 × 1015Hz

Substituting the values in the given expression of E:

E= (6.626 x 10-34) (3 x 1015)

=1.988 × 10–18J

(ii) Energy (E) of a photon having wavelength (λ) is given by the expression,

E = h(c/λ)

h = Planck’s constant = 6.626 × 10–34Js

c = velocity of light in vacuum = 3 × 108m/s

λ =0.50 Å = 0.50 x 10–10 m

Substituting the values in the given expression of E:

$E=\frac{\left(6.626×{10}^{-34}\right)\left(3×{10}^{8}\right)}{0.50×{10}^{-10}}=3.976×{10}^{-15}$ $E=3.98×{10}^{-15}J$

Q.114 For the reaction

2A(g) + B(g) → 2D(g)

∆Uθ = –10.5 kJ and ∆Sθ= –44.1 JK–1.

Calculate ∆Gθ for the reaction, and predict whether the reaction may occur spontaneously.

Ans.

For the given reaction,

2 A(g) + B(g) → 2D(g)

∆ng = 2 – (3)

= –1 mole

Substituting the value of ∆Uθ in the expression of ∆H:

∆Hθ = ∆Uθ + ∆ngRT

= (–10.5 kJ) – (–1) (8.314 × 10–3 kJ K–1 mol–1) (298 K)

= –10.5 kJ – 2.48 kJ

∆Hθ = –12.98 kJ

Substituting the values of ∆Hθ and ∆Sθ in the expression of ∆Gθ:

∆Gθ = ∆Hθ – T∆Sθ

= –12.98 kJ – (298 K) (–44.1 J K–1)

= –12.98 kJ + 13.14 kJ

∆Gθ = + 0.16 kJ

Since ∆Gθ for the reaction is positive, the reaction will not occur spontaneously.

Q.115 The equilibrium constant for a reaction is 10. What will be the value of ∆Gθ? R = 8.314

JK–1 mol–1, T = 300 K.

Ans.

From the expression,

∆Gθ = –2.303 RT logKeq

∆Gθ for the reaction,

= (2.303) × (8.314 JK–1 mol–1)× (300 K) × log10

= –5744.14 Jmol–1

= –5.744 kJ mol–1

Q.116

$\begin{array}{l}\text{Comment on the thermodynamic stability of NO}\left(\text{g}\right),\text{given}\\ \frac{1}{2}{N}_{2\left(g\right)}+\frac{1}{2}{O}_{2\left(g\right)}\to N{O}_{\left(g\right)}\text{};{\Delta }_{r}{H}^{\theta }=90\text{}kJ\text{}mo{l}^{-1}\\ \frac{1}{2}{N}_{\left(g\right)}+\frac{1}{2}{O}_{2\left(g\right)}\to N{O}_{2\left(g\right)}\text{};{\Delta }_{r}{H}^{\theta }=-74\text{}kJ\text{}mo{l}^{-1}\end{array}$

Ans.

The positive value of ∆rH indicates that heat is absorbed during the formation of NO(g).

This means that NO(g) has higher energy than the reactants (N2 and O2). Hence, NO(g) is unstable.

The negative value of ∆rH indicates that heat is evolved during the formation of NO2(g) from NO(g) and O2(g). The product, NO2(g) is stabilized with minimum energy. Hence, unstable NO(g) changes to unstable NO2(g).

Q.117 What would be the SI unit for the quantity pV2T2 /n?

Ans.

The SI unit for pressure, p is Nm–2.

The SI unit for volume, V is m3

The SI unit for temperature, T is K.

The SI unit for the number of moles, n is mol.

$\begin{array}{l}\text{Therefore, the SI unit of quantity}\frac{p{V}^{2}{T}^{2}}{n}\text{is given by}\\ \text{=}\frac{\left(N{m}^{-2}\right){\left({m}^{3}\right)}^{2}{\left(K\right)}^{2}}{mol}\\ ={\text{Nm}}^{4}{K}^{2}mo{l}^{-1}\end{array}$

Q.118 Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed under standard conditions. ∆fHθ = –286 kJ mol–1.

Ans.

It is given that 286 kJ mol–1 of heat is evolved on the formation of 1 mol of H2O(l). Thus, an equal amount of heat will be absorbed by the surroundings.

qsurr = +286 kJ mol–1

Entropy change (∆Ssurr) for the surroundings =

$\frac{{q}_{surr}}{T}$ $\frac{286{\text{kJ mol}}^{-1}}{298\text{k}}$

∆Ssurr =959.73 J mol -1 K-1

Q.119 In terms of Charles’ law explain why –273°C is the lowest possible temperature.

Ans.

Charles’ law states that at constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature.

At a given pressure the plot of volume vs. temperature is always a straight line for all gasses. If this line is extended to zero then it intersects at –273°C which is the lowest possible temperature.

Q.120 Critical temperature for carbon dioxide and methane are 31.1 °C and –81.9 °C respectively. Which of these has stronger intermolecular forces and why?

Ans.

The intermolecular forces of attraction are stronger in the case of CO2 because higher is the critical temperature of a gas, easier is its liquefaction. So, the intermolecular forces of attraction between the molecules of a gas are directly proportional to its critical temperature.

Q.121 Explain the physical significance of Van der Waals parameters.

Ans.

Physical significance of ‘a’:

Value of ‘a’ is measure of magnitude of intermolecular attractive forces within the gas and is independent of temperature and pressure.

Physical significance of ‘b’:

‘b’ is measure of the volume of a gas molecule.

Q.122 Calculate the wavelength, frequency and wave number of a light wave whose period is 2.0 × 10–10 s.

Ans.

Frequency (ν) of light = 1/period

$=\frac{1}{2.0×{10}^{-10}s}=5.0×{10}^{9}{s}^{-1}$

Wavelength (λ) of light = c/ν

Where, c = velocity of light in vacuum = 3×108 m/s

Substituting the value in the given expression of λ:

$\lambda =\frac{3×{10}^{8}}{5.0×{10}^{9}}=6.0×{10}^{-2}{m}^{}$

Wave number

$\left(\overline{v}\right)$

of light = 1/λ

$=\frac{1}{6.0×{10}^{-2}S}=1.66×{10}^{1}{m}^{-1}=16.66{m}^{-1}$

Q.123 What is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy?

Ans.

Energy (E) of a photon = hν

Energy (En) of ‘n’ photons = nhν

$n={E}_{n}\lambda /hc$

Where,

λ = wavelength of light = 4000 pm = 4000 ×10–12m

c = velocity of light in vacuum = 3 × 108 m/s

h = Planck’s constant = 6.626 × 10–34 Js

Substituting the values in the given expression of n, i.e.,

$n={E}_{n}\lambda /hc$

We get,

$n=\frac{{E}_{n}×\lambda }{h×c}=\frac{1J×4×{10}^{-9}m}{6.626×{10}^{-34}Js×3.0×{10}^{8}m{s}^{-1}}=2.012×{10}^{16}photons$

Hence, the number of photons with a wavelength of 4000 pm and energy of 1 J are 2.012 × 1016.

Q.124 A photon of wavelength 4 × 10–7m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate (i) the energy of the photon (eV), (ii) the kinetic energy of the emission, and (iii) the velocity of the photo electron (1 eV= 1.6020 × 10–19 J)

Ans.

(i) Energy (E) of a photon= hν= hc/λ

Where, h = Planck’s constant = 6.626 × 10–34 Js

c = velocity of light in vacuum = 3 × 108 m/s

λ = wavelength of photon = 4 × 10–7 m

Substituting the values in the given expression of E:

$E=\frac{\left(6.626×{10}^{-34}\right)\left(3×{10}^{8}\right)}{4×{10}^{-7}}=4.9695×{10}^{-19}J$

Hence, the energy of the photon is 4.97 × 10–19 J.

(ii) The kinetic energy of emission Ek is given by

= hν – hν0 = (E–W) eV

Where W is the work function. Let us place value of E and W in electron volt in the above equation, we get

${E}_{k}=\left(\frac{4.9695×{10}^{-19}}{1.6020×{10}^{-19}}\right)eV-2.13eV$

= (3.1020 – 2.13) eV

= 0.9720 eV

Hence, the kinetic energy of emission is 0.97 eV.

(iii) The velocity of a photoelectron (v) can be calculated by the following expression:

$\begin{array}{l}\frac{1}{2}m{v}^{2}=h\nu -h{\nu }_{0}\\ or\text{v}=\sqrt{\left\{2\left(h\nu -h{\nu }_{0}\right)/m\right\}}\end{array}$

Where, (hν – hν0) is kinetic energy of emission in Joules and ‘m’ is the mass of the photoelectron. Multiply kinetic energy with 1.6020 × 10–19 to convert into joules. Substitute these values in the given expression of v:

$\begin{array}{l}v=\sqrt{\frac{2×\left(0.9720×1.6020×{10}^{-19}\right)J}{9.10939×{10}^{-31}kg}}\\ =\sqrt{0.3418×{10}^{12}{m}^{2}{S}^{-2}}\end{array}$

v = 5.84 × 105 ms–1

Hence, the velocity of the photo electron is 5.84 × 105 ms-1.

Q.125 Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol–1

Ans.

Energy of sodium (E) = NAhcλ

$=\frac{\left({6.023510}^{23}mo{l}^{-1}\right)\left({6.626510}^{-34}Js\right)\left(3×{10}^{8}m{s}^{-1}\right)}{242×{10}^{-9}m}$

= 4.947 × 105 J mol–1

= 494 kJ mol-1

Q.126 A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57 mm. Calculate the rate of emission of quanta per second.

Ans.

Power of bulb, P = 25 Watt = 25 Js–1

Energy of one photon, E = hν = hc/λ

Substituting the values in the given expression of E:

$E=\frac{\left(6.626×{10}^{-34}Js\right)\left(3×{10}^{8}m{s}^{-1}\right)}{\left(0.57×{10}^{-6}m\right)}$

E = 34.87 × 10–20 J

Rate of emission of quanta per second, R

$=\frac{25}{34.87×{10}^{-20}}=7.169×{10}^{19}{s}^{-1}$

Q.127 Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency (ν0) and work function (W0) of the metal.

Ans.

Threshold wavelength of radian (λ0) =6800 Å

= 6800 × 10–10 m

Threshold frequency (ν0) of the metal= c/λ0

$=\frac{3×{10}^{8}m{s}^{-1}}{6.8×{10}^{-7}m}$

= 4.41 × 1014 s-1

Thus, the threshold frequency (ν0) of the metal is 4.41 × 1014 s–1

Hence, work function (W0) of the metal = hν0

$=\left(6.626×{10}^{-34}Js\right)\left(4.41×{10}^{14}{s}^{-1}\right)=2.922×{10}^{-19}J$

Q.128 What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2?

Ans.

Transition of hydrogen atom from an energy level with ni = 4 to nf = 2 will give rise to a spectral line of the Balmer series. The energy involved in the transition is given by the relation,

$\Delta E=2.18×{10}^{-18}J\left(\frac{1}{{n}_{i}^{2}}-\frac{1}{{n}_{f}^{2}}\right)=h\nu$

Substituting the values in the given expression of E:

$\Delta E=2.18×{10}^{-18}J\left(\frac{1}{{4}^{2}}-\frac{1}{{2}^{2}}\right)$ $\Delta E=2.18×{10}^{-18}J\left(\frac{\left(1-4\right)}{16}\right)$ $\Delta E=2.18×{10}^{-18}J\left(\frac{-3}{16}\right)$

E = – (4.0875 × 10–19J)

Wavelength of light emitted, λ = hc/E

Substitute these values in the given expression of λ:

$\lambda =\frac{\left(6.626×{10}^{-34}Js\right)\left(3×{10}^{8}m{s}^{-1}\right)}{\left(4.0875×{10}^{-19}\right)}$ $\lambda =486.3×{10}^{-9}m=486\text{}nm$

Q.129 How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of H atom (energy required to remove the electron from n = 1 orbit).

Ans.

The expression of energy is given by,

${E}_{n}=\frac{-\left(2.18×{10}^{-18}\right){Z}^{2}}{{n}^{2}}$

Z = atomic number of the atom

n = principal quantum number

For ionisation from n1= 5 to n2 = ∞

$\Delta E={E}_{\propto }-{E}_{5}$ $=\left(\left\{\frac{-\left(2.18×{10}^{-18}J\right)\left({1}^{2}\right)}{{\left(\propto \right)}^{2}}\right\}-\left\{\frac{-\left(2.18×{10}^{-18}J\right)\left({1}^{2}\right)}{{5}^{2}}\right\}\right)$

$=\left(2.18×{10}^{-18}J\right)\left(\frac{1}{\left({5}^{2}\right)}\right)\text{}Since,\text{}\frac{1}{\propto }=0$

= 0.0872 x 10-18 J

ΔE = 8.72 x 10-20 J

Hence, the energy required for ionization from n = 5 to n = ∞ is 8.72 × 10–20J.

Energy required for n1 = 1 to n = ∞,

ΔE’ = E – E1

$=\left(\left\{\frac{-\left(2.18×{10}^{-18}J\right)\left({1}^{2}\right)}{{\left(\propto \right)}^{2}}\right\}-\left\{\frac{-\left(2.18×{10}^{-18}J\right)\left({1}^{2}\right)}{{1}^{2}}\right\}\right)$ $=\left(2.18×{10}^{-18}J\right)\left(1-0\right)$

= 2.18 x 10-18 J

On comparing the ionisation energies,

ΔE/ΔE’ =2.18 x 10-18 J / 8.72 x 10-20 J = 25

Thus, the energy required to remove an electron from n=1 orbit in a hydrogen atom is 25 times the energy required to remove an electron from n = 5 orbit.

Q.130 What is the maximum number of emission lines when the excited electron of an H atom in n = 6 drops to the ground state?

Ans.

When the excited electron of an H atom in n = 6 drops to the ground state, the following transitions are possible

Thus, a total number of (5 + 4 + 3 + 2 + 1) 15 lines will be obtained in the emission spectrum.

The numbers of spectral lines produced when an electron in the nth level drops down to the ground state is given by n (n – 1)/2.

When n = 6, the Number of spectral lines = 6 (6 – 1)/2 = 15

Q.131 (i) The energy associated with the first orbit in the hydrogen atom is –2.18 ×10–18 J atom–1. What is the energy associated with the fifth orbit?

(ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom.

Ans.

(i) Energy associated with the fifth orbit of hydrogen atom is calculated as:

${E}_{5}=\frac{-\left(2.18×{10}^{-18}\right)}{{5}^{2}}=\frac{-2.18×{10}^{-18}}{25}$

E5 = – 8.72 × 10–20 J

(ii) Radius of Bohr’s nth orbit for hydrogen atom is given by,

rn = (0.0529 nm) n2

For n = 5, let us calculate radius for hydrogen atom for fifth orbit

r5 = (0.0529 nm) (5)2

r5 = 1.3225 nm

Q.132 Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.

Ans.

For the Balmer series, ni = 2. Thus, the expression of wave number

$\left(\overline{v}\right)$

is given by

$\overline{v}=\left(\frac{1}{{\left(2\right)}^{2}}-\frac{1}{{n}_{f}^{2}}\right)\left(1.097×{10}^{7}\right){m}^{-1}$ $\overline{v}=1/\lambda$

Wave number is inversely proportional to wavelength of transition. Hence, the longest wavelength transition, wave number has to be the smallest.

For wave number to be minimum, nf should be minimum. For the Balmer series, a transition from ni = 2 to nf = 3 is allowed. Hence, by taking nf = 3, we get:

$\begin{array}{l}\overline{v}=\left(1.097×{10}^{7}\right)\left(\frac{1}{{\left(2\right)}^{2}}-\frac{1}{{\left(3\right)}^{2}}\right)\\ \overline{v}=\left(1.097×{10}^{7}\right)\left(\frac{1}{4}-\frac{1}{9}\right)\\ \overline{v}=\left(1.097×{10}^{7}\right)\left(\frac{9-4}{36}\right)\end{array}$

ν = 1.5236 × 106 m–1

Q.133 What is the energy in Joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is –2.18 × 10–11 ergs.

Ans.

Let us convert ground state electron energy into joules.

Ground state electron energy = – 2.18 × 10–11 ergs

= – 2.18 × 10–11 × 10–7 J

= – 2.18 × 10–18 J

Energy (E) of the nth Bohr orbit of an atom is given by,

${E}_{n}=\frac{-\left(2.18×{10}^{-18}\right){Z}^{2}}{{n}^{2}}$

Where, Z = atomic number of the atom

Energy required for shifting the electron from n = 1 to n = 5 is given as:

ΔE = E5 -E1

$=\frac{-\left(2.18×{10}^{-18}\right){\left(1\right)}^{2}}{{5}^{2}}-\left(2.18×{10}^{-18}\right)$ $=\left(2.18×{10}^{-18}\right)\left(1-\frac{1}{25}\right)$ $=\left(2.18×{10}^{-18}\right)\left(\frac{24}{25}\right)$

= 2.0928 x 10-18 J

Wave length of emitted light = hc/E

$=\frac{\left(6.626×{10}^{-34}\right)\left(3×{10}^{8}\right)}{\left(2.0928×{10}^{-18}\right)}$

= 9.498 x 10-8 m

Q.134 The electron energy in hydrogen atom is given by En = (–2.18 × 10–18)/n2 J. Calculate the energy required to remove an electron completely from the n= 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?

Ans.

Given,

${E}_{n}=-\frac{2.18×{10}^{-18}}{{n}^{2}}J$

Energy required for ionization from n = 2 is given by,

= 0.545 x 10-18 J

$\Delta E={E}_{\propto }-{E}_{2}$ $=\left(\left\{\frac{-\left(2.18×{10}^{-18}J\right)}{{\left(\propto \right)}^{2}}\right\}-\left\{\frac{-\left(2.18×{10}^{-18}J\right)}{{2}^{2}}\right\}\right)$ $=\left(\frac{2.18×{10}^{-18}}{4}-0\right)$

ΔE = 5.45 x 10-19 J

λ = hc/ΔE

Here, λ is the longest wavelength causing the transition.

$\lambda =\frac{\left(6.626×{10}^{-34}\right)\left(3×{10}^{8}\right)}{\left(5.45×{10}^{-19}\right)}=3.647×{10}^{-7}m$

= 3647 x 10-10 m

= 3647 Å

Q.135 Calculate the wavelength of an electron moving with a velocity of 2.05 ×107 ms–1.

Ans.

According to de Broglie’s equation,

λ = h/mv

Where,

λ = wavelength of moving particle

m = mass of particle

v= velocity of particle

h = Planck’s constant

Substituting the values in the expression of λ:

$\begin{array}{l}\lambda =\frac{6.626×{10}^{-34}Js}{\left(9.10939×{10}^{-31}kg\right)\left(2.05×{10}^{7}m{s}^{-1}\right)}\\ \lambda =3.548×{10}^{-11}m\end{array}$

Hence, the wavelength of the electron moving with a velocity of 2.05 × 107 ms–1 is 3.548 × 10–11 m.

Q.136 The mass of an electron is 9.1 × 10–31 kg. If its K.E. is 3.0 × 10–25 J, calculate its wavelength.

Ans.

From de Broglie’s equation,

λ = h/mv

Given,

Kinetic energy (K.E.) of the electron = 3.0 × 10–25 J.

K.E = ½ mv2

$\begin{array}{l}Velocity\left(v\right)=\sqrt{2K.E./m}\\ =\sqrt{\frac{2\left(3.0×{10}^{25}J\right)}{9.10939×{10}^{-31}kg}}\end{array}$

v = 812 ms-1

Substituting the value in the expression of λ:

$\lambda =\frac{6.626×{10}^{-34}Js}{\left(9.10939×{10}^{-31}kg\right)\left(812m{s}^{-1}\right)}$

λ= 8.9627 x 10-7 m

Hence, the wavelength of the electron is 8.9627 × 10–7 m.

Q.137 Which of the following are isoelectronic species, i.e., those having the same number of electrons? Na+, K+, Mg2+, Ca2+, S2–, Ar

Ans.

Number of electrons in Na+ = 11–1 = 10

K+ = 19–1 = 18

Mg2+= 12–2 = 10

Ca2+ = 20–2 = 18

S2– = 16+2 = 18

Ar = 18

Hence Na+ and Mg2+ are isoelectronic species and

K+, Ca2+, S2–, Ar are isoelectronic species.

Q.138 (i)Write the electronic configurations of the following ions: (a) H (b) Na+(c) O2–(d) F

(ii) What are the atomic numbers of elements whose outermost electrons are represented by

(a) 3s1 (b) 2p3 and (c) 3p5?

(iii)Which atoms are indicated by the following configurations?

(a) [He] 2s1

(b) [Ne] 3s23p3

(c) [Ar] 4s23d1.

Ans.

(i) (a) H ion:

The electronic configuration of H atom is 1s1.

A negative charge on hydrogen indicates the gain of electron by it.

Electronic configuration of H = 1s2

(b) Na+ ion:

The electronic configuration of Na atom is 1s22s22p63s1.

A positive charge on the species indicates the loss of an electron by it.

Electronic configuration of Na+ =1s2 2s2 2p6 3s0

=1s2 2s2 2p6

(c)O2– ion:

The electronic configuration of O atom is 1s2 2s2 2p4

A dinegative charge on the species indicates that two electrons are gained by it.

Electronic configuration of O2–ion = 1s2 2s2 2p6

(d)Fion:

The electronic configuration of F atom is 1s2 2s2 2p5.

A negative charge on the species indicates the gain of an electron by it.

Electron configuration of Fion = 1s2 2s2 2p6

(ii) (a) 3s1:

Completing the electron configuration of the element as = 1s2 2s2 2p6 3s1

Number of electrons present in the atom of the element = 2 + 2 + 6 + 1 = 11

Atomic number of the element = 11

(b)2p3:

Completing the electron configuration of the element as

= 1s22s22p3

Number of electrons present in the atom of the element = 2 + 2 + 3 = 7

Atomic number of the element = 7

(c) 3p5:

Completing the electron configuration of the element as

= 1s22s22p5

Number of electrons present in the atom of the element = 2 + 2 + 5 = 9

Atomic number of the element = 9

(iii) (a)[He] 2s1:

The electronic configuration of the element is

[He] 2s1= 1s22s1.

Atomic number of the element = 3

The element is lithium (Li).

b)[Ne] 3s23p3 :

The electronic configuration of the element is

[Ne]3s23p3 = 1s22s22p63s23p3

Atomic number of the element = 15

Hence, the element with the electronic configuration

[Ne] 3s23p3 is phosphorus (P).

(c)[Ar] 4s23d1:

The electronic configuration of the element is [Ar]4s23d1= 1s22s22p63s23p64s23d1.

Atomic number of the element = 21

Hence, the element with the electronic configuration

[Ar] 4s23d1 is scandium (Sc).

Q.139 What is the lowest value of n that allows g- orbitals to exist?

Ans.

For g-orbital, l (Azimuthal quantum number) = 4,

As for any value of principal quantum number(n), the Azimuthal quantum number (l) can have a value from zero to (n–1).

∴ For l= 4, minimum value of n= 5

Q.140 An electron is in one of the 3d orbitals. Give the possible values of n, l and ml for this electron.

Ans.

For the 3d orbital:

Principal quantum number (n) = 3

Azimuthal quantum number (l) = 2

Magnetic quantum number (ml) = –2,–1, 0, 1, 2

Q.141 An atom of an element contains 29 electrons and 35 neutrons. Deduce

(i) the number of protons and (ii) the electronic configuration of the element.

Ans.

(i) For a neutral atom, number of protons is equal to the number of electrons. Thus the number of protons in the atom of given element = 29.

(ii)The electronic configuration of the atom is

= 1s22s22p63s23p64s13d10.

Q.142 Give the number of electrons in the species H2+, H2, and O2+.

Ans.

Number of electrons present in hydrogen molecule (H2)

=1+1 = 2

Number of electrons in H2+ = 2-1= 1

Number of electrons in H2= 1 + 1 = 2

Number of electrons present in oxygen molecule (O2) = 8 + 8 = 16

Number of electrons in O2+ = 16 – 1 = 15

Q.143 (i) An atomic orbital has n = 3. What are the possible values of l and ml?

(ii) List the quantum numbers (ml and l) of electrons for 3d orbital.

(iii) Which of the following orbitals are possible? 1p, 2s, 2p and 3f

Ans.

(i) n = 3 (Given)

For a given value of n, l can have values from 0 to

(n– 1).

For n= 3, l= 0, 1, 2

For a given value of l, ml can have (2l+ 1) values

For l= 0, m= 0

l= 1, m= – 1, 0, 1

l= 2, m= – 2, – 1, 0, 1, 2

Thus, for n= 3 l= 0, 1, 2 m0= 0

m1= – 1, 0, 1 m2= – 2, – 1, 0, 1, 2

(ii) For 3d orbital, l= 2. For a given value of l, ml can have (2l+ 1) values i.e., 5 values.

∴ For l = 2, ml = – 2, – 1, 0, 1, 2

(iii) Among the given orbitals only 2s and 2p are possible. 1p and 3f cannot exist.

For p-orbital, l= 1.

For a given value of n, l can have values from zero to (n– 1).

∴ For l =1, the minimum value of n is 2.

Similarly, for f-orbital, l = 3.

For l= 3, the minimum value of n is 4.

Hence, 1p and 3f do not exist.

Q.144 Using s, p, d notations, describe the orbital with the following quantum numbers. (a) n= 1, l = 0; (b)n= 3; l=1 (c)n = 4; l= 2; (d)n= 4; l=3

Ans.

(a) For n= 1, l= 0 (Given), the orbital is 1s.

(b)For n= 3 and l= 1, the orbital is 3p.

(c) For n= 4 and l= 2, the orbital is 4d.

(d) For n= 4 and l= 3, the orbital is 4f.

Q.145 Explain, giving reasons, which of the following sets of quantum numbers are not possible.

$\begin{array}{l}a\right)\text{}n=0\text{}l=0\text{}{m}_{l}=0\text{}{m}_{s}=+1/2\\ b\right)\text{}n=1\text{}l=0\text{}{m}_{l}=0\text{}{m}_{s}=-1/2\\ c\right)\text{}n=1\text{}l=1\text{}{m}_{l}=0\text{}{m}_{s}=+1/2\\ d\right)\text{}n=2\text{}l=1\text{}{m}_{l}=0\text{}{m}_{s}=-1/2\\ e\right)\text{}n=3\text{}l=3\text{}{m}_{l}=-3\text{}{m}_{s}=+1/2\\ f\right)\text{}n=3\text{}l=1\text{}{m}_{l}=0\text{}{m}_{s}=+1/2\end{array}$

Ans.

(a) The given set of quantum numbers is not possible because the value of the principal quantum number (n) cannot be zero.

(b) The given set of quantum numbers is possible.

(c)The given set of quantum numbers is not possible.

For a given value of n, ‘l’ can have values from zero to (n– 1). For n= 1, l= 0 and not 1.

(d)The given set of quantum numbers is possible.

(e) The given set of quantum numbers is not possible.

For n= 3, l= 0 to (3 – 1),i.e., l= 0 to 2 or 0, 1, 2

(f)The given set of quantum numbers is possible.

Q.146 How many electrons in an atom may have the following quantum numbers?

(a) n= 4, ms=–1/2 and (b) n=3, l=0

Ans.

(a) Total number of electrons in an atom for a value of n= 2n2

For n=4, total number of electrons= 2(4)2 =32.

The given element has a fully filled orbital as

1s2 2s2 2p6 3s2 3p6 4s2 3d10

Hence, all the electrons are paired.

Number of electrons (having n= 4 and ms=–1/2) = 16

(b) n = 3, l= 0 indicates that the electrons are present in the 3s orbital. Therefore, the number of electrons having n= 3 and l= 0 is 2.

Q.147 Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.

Ans.

Since a hydrogen atom has only one electron, according to Bohr’s postulate, the angular momentum of electron is given by:

$mvr=nh/2\pi \text{}\dots \dots \dots \dots \dots \dots \dots \dots \dots \left(1\right)$

Where,

n = 1, 2, 3…

According to de Broglie’s equation:

$\begin{array}{l}\lambda =h/mv\\ Or,\text{mv}=h/\lambda \text{}\dots \dots \dots \dots \dots \dots \dots \dots \dots \left(2\right)\end{array}$

Substituting the value of ‘mv’ from equation (2) in equation (1):

$\begin{array}{l}hr/\lambda =nh/2\pi \\ Or,\text{2}\pi \text{r}=n\lambda \text{}\dots \dots \dots \dots \dots \dots \dots \dots \dots \left(3\right)\end{array}$

Since ‘2πr’ represents the circumference of the Bohr orbit (r), it is proved by equation (3) that the circumference of the Bohr orbit of the hydrogen atom is an integral multiple of de Broglie’s wavelength associated with the electron revolving around the orbit.

Q.148 What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n= 4 to n= 2 of He+ spectrum?

Ans.

For He+ ion, expression of the wave number

$\left(\overline{v}\right)$

associated with the Balmer transition, n=4 to n=2

Where, n1= 2, and n2=4

Z = atomic number of helium

$\overline{v}=\frac{1}{\lambda }=R{Z}^{2}\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)$ $=4R\frac{\left(4-1\right)}{16}$ $\begin{array}{l}\overline{v}=1/\lambda =3R/4\text{}\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \left(2\right)\\ \lambda =4/3R\end{array}$

Comparing above two equations the desired transition for hydrogen will have the same wavelength as that of He+.

$\begin{array}{l}=R{\left(1\right)}^{2}\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)=\frac{3R}{4}\\ =\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)=\frac{3}{4}\text{}\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \left(1\right)\end{array}$

By hit and trail method, the equality given by equation (1) is true only when

n1= 1 and n2= 2.

The transition for n2= 2 to n1= 1 in hydrogen spectrum would have the same wavelength as Balmer transition n= 4 to n = 2 of He+ spectrum.

Q.149 Calculate the energy required for the process

$H{e}^{+}\left(g\right)\to H{e}^{2+}\left(g\right)+{e}^{-}$

The ionization energy for the H atom in the ground state is 2.18 × 10–18 J atom–1.

Ans.

Energy associated with hydrogen-like species is given by,

${E}_{n}=-2.18×{10}^{-18}\left({Z}^{2}/{n}^{2}\right)J$

For ground state of hydrogen atom,

ΔE = E–E1

$=0-\left[-2.18×{10}^{-18}\left\{\frac{{\left(2\right)}^{2}}{{\left(1\right)}^{2}}\right\}\right]J$

ΔE= 2.18x 10-18 J

For the given process,

$H{e}^{+}\left(g\right)\to H{e}^{2+}\left(g\right)+{e}^{-}$

An electron is removed from n= 1 to n= ∞.

ΔE = E-E1

$=0-\left[-2.18×{10}^{-18}\left\{\frac{{\left(2\right)}^{2}}{{\left(1\right)}^{2}}\right\}\right]J$

ΔE= 8.72 x 10-18 J

The energy required for the process= 8.72 x 10-18 J.

Q.150 If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across length of scale of length 20 cm long.

Ans.

1 m= 100 cm, or 1cm= 10-2 m

Length of the scale = 20 cm= 20 × 10–2m

Diameter of a carbon atom = 0.15 nm= 0.15 × 10–9m

One carbon atom occupies 0.15 × 10–9m

Number of carbon atoms that can be placed in a straight line

$=\frac{20×{10}^{-2}m}{0.15×{10}^{-9}m}=133.33×{10}^{7}$

= 1.33 x 109

Q.151 2 × 108atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm.

Ans.

Number of carbon atoms present = 2 × 108
Length of the given arrangement = 2.4 cm
Diameter of carbon atom

$=\frac{2.4×{10}^{-2}m}{2×{10}^{8}m}=1.2×{10}^{-10}$

Radius of carbon atom = (Diameter/2)

$=\frac{1.2×{10}^{-10}m}{2}=6.0×{10}^{-11}m$

Q.152 The diameter of zinc atom is 2.6Å Calculate (a) radius of zinc atom in pm and (b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise.

Ans.

(a) Radius of zinc = (Diameter/2)

=2.6/2 Å

$\begin{array}{l}=1.3×{10}^{-10}m\\ =130×{10}^{-12}m=130\text{}pm\end{array}$

(b) Length of the arrangement = 1.6 cm

= 1.6 x 10-2 m Diameter of zinc atom = 2.6 × 10–10m

Number of zinc atoms present in the arrangement

$\begin{array}{l}=\frac{1.6×{10}^{-2}m}{2.6×{10}^{-10}m}\\ =0.6153×{10}^{8}m\\ =6.153×{10}^{7}m\end{array}$

Q.153 A certain particle carries 2.5 × 10–16C of static electric charge. Calculate the number of electrons present in it.

Ans.

Charge on one electron = 1.6022 × 10–19 C

= 1.6022 ×10–19 C charge is carried by 1 electron

Number of electrons carrying a charge of 2.5 × 10–16 C

$=\frac{2.5×{10}^{-16}C}{1.6022×{10}^{-19}C}=1.560×{10}^{3}C=1560C$

Q.154 In Milikan’s experiment, static electric charge on the oil drops has been obtained by shining X-rays. If the static electric charge on the oil drop is –1.282 × 10–18C, calculate the number of electrons present on it.

Ans.

Charge on the oil drop = – 1.282 ×10–18 C

Charge on one electron = – 1.6022 × 10–19 C

∴Number of electrons present on the oil drop

$\begin{array}{l}=\frac{-1.282×{10}^{-18}C}{-1.6022×{10}^{-19}C}\\ =0.8001×{10}^{1}=8.0\end{array}$

Q.155 In Rutherford’s experiment, generally the thin foil of heavy atoms, like gold, platinum etc. have been used to be bombarded by the α-particles. If the thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results?

Ans.

A thin foil of lighter atoms cannot give the same results as given with the foil of heavier atoms.

The lighter atoms will not cause enough deflection of α- particles because they would be able to carry very little positive charge.

Q.156 Symbols

${}_{35}^{79}Br\text{and}{\text{}}^{79}Br$

can be written, whereas symbols

${}_{79}^{35}B\mathrm{r and}\text{}{\text{}}^{35}Br$

Ans.

The convenient way of representing an element with its atomic mass (A) and atomic number (Z) is

${}_{Z}^{A}X$

Atomic mass of Br is 79, but atomic number Z is 35, thus the symbol

${}_{35}^{79}Br$

is acceptable but

${}_{79}^{35}Br$

is not acceptable.

Again, the atomic number of an element is constant, but the atomic mass of an element depends

upon the relative abundance of its isotopes, so

${}^{79}Br$

can be written but

${}^{35}Br$

can not be written.

Q.157 An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.

Ans.

Let the number of protons in the element be x. Number of neutrons in the element = x + 31.7% of x
= x + 0.317 x = 1.317 x
According to the question,
Mass number of the element = 81
(Number of protons + number of neutrons) = 81

$X=\frac{81}{2.317}=34.95$

X= 35
The number of protons in the element is 35.
Since the atomic number of an atom is defined as the number of protons present in its nucleus, the atomic number of the given element is 35.
The atomic symbol of the element is

${}_{35}^{81}Br$

Q.158 An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more neutrons than the electrons, find the symbol of the ion.

Ans.

Let the number of electrons in the ion carrying a negative charge be x.

Then, number of neutrons present = x+ 11.1% of x

= x + 0.111x = 1.111x

Number of electrons in the neutral atom = (x– 1)

(When an ion carries a negative charge, it carries an extra electron).

Number of protons in the neutral atom = (x – 1)

Mass number of the ion = 37

So, (x– 1) + 1.111x= 37

Or, 2.111x= 38 or, x= 18

The symbol of the ion is

${}_{17}^{37}C{l}^{-}$

Q.159 An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol to this ion.

Ans.

Let the number of electrons present A3+ be x.
Number of neutrons in it = x + 30.4% of x = 1.304x
Since the ion is tri-positive,
Number of electrons in neutral atom = x+ 3 = Number of protons in neutral atom
Mass number of the ion = 56

$\begin{array}{l}\left(x+3\right)+\left(1.304x\right)=56\\ or,\text{}2.304x=53\\ X=\frac{53}{2.304}\text{}\text{}or,\text{}X=23\end{array}$

Number of protons = x + 3 = 23 + 3 = 26
The symbol of the

${}_{26}^{56}F{e}^{3+}$

Q.160 Arrange the following type of radiations in increasing order of frequency: (a) radiation from microwave oven (b) amber light from traffic signal (c) radiation from FM radio (d) cosmic rays from outer space and (e) X-rays.

Ans.

The increasing order of frequency is as follows:

Radiation from FM radio < radiation from microwave oven < amber light < X- rays < Cosmic rays

Frequency is inversely proportional to the wavelength.

The increasing order of wavelength is as follows:

Cosmic rays < X-rays < amber light < radiation from microwave ovens < radiation of FM radio.

Q.161 Nitrogen laser produces a radiation at a wave length of 337.1 nm. If the number of photons emitted is 5.6 × 1024, calculate the power of this laser.

Ans.

Power of laser = Energy with which it emits photons

$=E=\frac{Nhc}{\lambda }$

Where,

N= number of photons emitted

h = Planck’s constant

Substituting the values in the given expression of Energy (E):

$E=\frac{\left(5.4×{10}^{24}\right)\left(6.626×{10}^{-34}Js\right)\left(3×{10}^{8}m{s}^{-1}\right)}{337.1×{10}^{-9}m}$

Energy (E):
E = 0.32 × 107J
= 3.2 × 106J
The power of the laser is 3.2 × 106J.

Q.162 Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, calculate (a) the frequency of emission, (b) distance traveled by this radiation in 30 s (c) energy of quantum and (d) number of quanta present if it produces 2 J of energy.

Ans.

Wavelength of radiation emitted = 616 nm

= 616 × 10–9m (Given)

$\nu =c/\lambda$

Where,

Substituting the values in the given expression (ν)

$\begin{array}{l}\nu =\frac{3.0×{10}^{8}m/s}{616×{10}^{-9}m}\\ =4.87×{10}^{8}×{10}^{9}×{10}^{-3}{s}^{-1}\end{array}$

= 4.87 × 1014s–1

Frequency of emission (ν) = 4.87 × 1014s-1

$\left(b\right)\text{Velocity of radiation},\text{}\left(\text{c}\right)\text{}=\text{3}.0\text{}×\text{1}{0}^{\text{8}}{\text{ms}}^{–\text{1}}$

Distance travelled by this radiation in 30 s
= (3.0 × 108ms–1) (30 s)
= 9.0 × 109m
(6.626 × 10–34Js) (4.87 × 1014s–1)
(c) Energy of quantum (E) = 32.27 × 10–20J

=

$\begin{array}{l}\text{}\\ \left(d\right)\text{Energy of one photon}\left(\text{quantum}\right)\text{}=\text{32}.\text{27}×\text{1}{0}^{–\text{2}0}\text{J}\end{array}$