# NCERT Solutions Class 11 Chemistry Chapter 1

**NCERT Solutions for Class 11 Chemistry Chapter 1 – Some Basic Concepts of Chemistry**

Extramarks Chemistry class 11 chapter 1 NCERT solutions were created keeping the needs of Class 11 students in mind by chemistry subject matter experts. The NCERT Solutions on this page provide step-by-step explanations to assist students in answering comparable questions that may appear on their term – I examinations.

NCERT Solutions for Class 11 Chemistry Chapter 1 have been developed by qualified teachers in a simple and clear language in compliance with the most recent CBSE Syllabus 2022-23 and its guidelines. . The answers cover the fundamental ideas in order to provide a high-quality learning experience for Class 11 students. To complete a portion of the term – I syllabus before the first term test, you can get the Class 11 Chemistry NCERT Solutions Chapter 1 from Extramarks’ website.

Our chemistry experts explain the answers to all the questions using the latest NCERT syllabus following the CBSE pattern. Some essential topics in Chemistry Class 11 NCERT Solutions are provided to make your learning easier and more interesting.

** NCERT Solutions for Class 11 Chemistry Chapter 1- Some Basic Concepts of Chemistry**

**CBSE Class 11 Chemistry Chapter 1: Chapter Summary **

Chemistry is crucial because it affects every aspect of life. Chemistry subjects have the qualities and structure of substances, as well as the changes that they undergo. Matter is found in all substances and can exist in three states: solid, liquid, or gas. Students will learn about the nature of materials and chemical rules of combination in this chapter. This chapter will assist students in establishing a solid foundation in Chemistry. Students will learn about atomic and molecular masses, as well as the mole idea, chemical reactions, and molecular formulas.

As a result, the chapter’s practical problems will aid students in better understanding the topics. The Law of Conservation of Mass, the Law of Definite Proportions, the Law of Multiple Proportions, the International System of Units, Weights and Measures, and other concepts will be covered. This is one of the foundational chapters in Chemistry, but it doesn’t carry much weightage. Make sure you understand the whole chapter before answering the questions.

Important Topics: Mole Fraction, Stoichiometry, Equivalent Mass and Normality, Atomic Number, Mass number and Isotypes, Avogadro’s Law and Percent Composition.

**NCERT Solutions for Class 11 Chemistry Chapter 1 – A Quick glance through the Chapter**

**Atomic Mass and Molecular Mass**

One mass unit is defined as a mass that is one-twelfth the mass of one carbon–12 atoms. The Molecular Mass is the sum of the atomic masses of the atoms in a molecule. You may get it by multiplying each element’s mass by the number of atoms in each of its atoms and adding the results. Gram molecular mass refers to molecular mass expressed in grams. The sum of all atoms’ atomic masses during a formula unit of the compound is known as the formula mass. The mole is the amount of substance in a system that includes the same number of elementary entities as there are atoms in 0.012 kilograms of carbon-12; it is denoted by the sign -mol. This number of entities per mol is so important that it has its own name and symbol. In honour of Avogadro, it is known as the ‘Avogadro constant.’

**Concepts of Mass, Weight, Volume and Temperature**

Mass, which is a constant quantity, is the matter present in a substance. The kilogram is the SI unit for mass (Kg). Weight, on the other hand, is the force exerted on a substance by gravity. Due to variations in gravity, it varies from location to place. The SI unit of weight is the Newton. Again, volume refers to the amount of space occupied by a body. In addition, the cubic metre is a SI unit. A litre is another common measurement. One cubic metre equals 1000 litres. The density, which has the SI unit Kg per cubic metre or gram per cubic metre, is the quantity of mass per unit volume. The measurement of heat in a substance is called temperature. SI unit is Kelvin, and other standard units are Degree Celsius (°C) and degree. Fahrenheit (°F).

**How can our subject experts help you to improve yourself?**

Along with NCERT Solutions Class 11 Chemistry Chapter 1 of all topics, Extramarks also offers past years’ question papers to assist you to understand the question pattern. The question papers come with answer keys that are prepared by our qualified faculty experts with years of experience. Students can also attend **LIVE online sessions **for one-on-one interaction, and to clarify their doubts related to the topic.

**Why Should You Prefer NCERT Solutions prepared by Extramarks for Class 11 Chemistry Chapter 1? Some Basic Concepts of Chemistry prepared by the subject Experts of Extramarks?**

The large number of new concepts discussed make Chapter 1 difficult to comprehend for students. Extramarks offers Chapter 1 NCERT Solutions for Class 11 Chemistry to explain the concepts in a well-organised, comprehensible, and straightforward manner. It follows the latest syllabus to help you get better grades in tests. The solutions are prepared by Extramarks subject-matter experts who have years of experience in teaching, which itself speaks of the credibility and accuracy of the learning material.

The NCERT Solutions presented here have been created with one goal in mind: to assist students in preparing for their term-by-term examinations and passing them with flying colours. The NCERT Solutions comprise all of the exercise questions from the NCERT textbook, thus they cover a lot of significant questions that might come in the term I and II exams.

**Q.1 ****Calculate the molecular mass of the following:**

**1: H _{2}O**

**2: CO _{2}**

**3: CH _{4}**

**Ans.**

1. Molecular mass of H_{2}O is:

= (2 × Atomic Mass of Hydrogen) + (2 × Atomic Mass of Oxygen)

= [2(1.0084) + 1(16.00 u)]

= 2.0168 u + 16.00 u

= 2.016 u + 16.00 u

= 18.0168 u

2. Molecular mass of CO_{2} is:

= (1 × Atomic mass of carbon) + (2 × Atomic mass of oxygen)

= [1(12.011 u) + 2 (16.00 u)]

= 12.011 u + 32.00 u

= 44.011

3. The molecular mass of CH_{4}

= (1 × Atomic mass of carbon) + (4 × Atomic mass of hydrogen)

= [1(12.011 u) + 4 (1.008 u)]

= 12.011 u + 4.032 u

= 16.043 u

**Q.2 ****Calculate the amount of carbon dioxide that could be produced when**

**(i) 1 mole of carbon is burnt in air.**

**(ii) 1 mole of carbon is burnt in 16 g of dioxygen.**

**(iii) 2 moles of carbon are burnt in 16 g of dioxygen.**

**Ans.**

The balanced reaction of combustion of carbon can be written

(i) As per the balanced equation, 1 mole of carbon burns in1 mole of dioxygen (air) to produce 1 mole of carbon dioxide.

(ii) According to the question, only 16 g of dioxygen is available. Hence, it will react with 0.5 mole of carbon to give 22 g of carbon dioxide. Hence, it is a limiting reactant.

(iii) According to the question, only 16 g of dioxygen is available. It is a limiting reactant. Thus, 16 g of dioxygen can combine with only 0.5 mole of carbon to give 22 g of carbon dioxide.

**Q.3 ****Calculate the mass percent of different elements present in sodium sulphate (Na _{2}SO_{4}).**

**Ans.**

The molecular formula of sodium sulphate is: Na_{2}SO_{4}

Molar mass of Na_{2}SO_{4}= [(2 × 23.0) + (32.066) + 4 (16.00)] = 142.066 g

\begin{array}{l}Masspercentofanelement=\frac{Massofthatelementinthecompound}{Molarmassofthecompound}x100\\ \therefore Masspercentofsodium:\\ Masspercentofsodium=\frac{46.0g}{142.066g}x100\\ =\text{32}.\text{379}\\ =\text{32}.\text{4}\%\\ Masspercentofsulphur=\frac{32.066g}{142.066g}x100\\ =22.57\\ =22.6\%\\ Masspercentofoxygen=\frac{64.0g}{142.066g}x100\\ =\text{45}.0\text{49}\\ =\text{45}.0\text{5}\%\end{array}

**Q.4 ****Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.**

**Ans.**

% of iron by mass=69.9% [Given]

% of oxygen by mass=30.1% [Given]

\begin{array}{l}Relativemolesofironinironoxide=\frac{\%ofironbyMass}{AtomicMassofIron}\\ =\frac{69.9}{55.85}\\ =1.25\\ Relativemolesofironinironoxide=\frac{\%ofironbyMass}{AtomicMassofOxygen}\\ \frac{\%ofOxygenbyMass}{AtomicMassofOxygen}\\ =\frac{30.1}{16.00}\\ =1.88\end{array}

Simplest molar ratio of iron to oxygen:

= 1.25: 1.88

= 1: 1.5

= 2: 3

∴The empirical formula of the iron oxide is Fe_{2 }O_{3}

**Q.5 ****Calculate the mass of sodium acetate (CH _{3 }COONa) required to make 500 ml of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol^{-1}**

**Ans.**

0.375 M aqueous solution of sodium acetate

≡ 1000 mL of solution containing 0.375 moles of sodium acetate

∴ Number of moles of sodium acetate in 500 mL

=\frac{0.375}{1000}x500

= 0.1875 mole

Molar mass of sodium acetate = 82.0245 g mol^{-1} (Given)

∴ Required mass of sodium acetate = (82.0245 g mol^{-1}) (0.1875 mole)

= 15.38 g

**Q.6 ****Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL ^{-1} and the mass per cent of nitric acid in it being 69%.**

**Ans.**

Mass percent of nitric acid in the sample = 69 % [Given]

Thus, 100 g of nitric acid contains 69 g of nitric acid by mass.

Molar mass of nitric acid (HNO)_{3}

= {1 + 14 + 3(16)} g mol^{–1}

= 1 + 14 + 48

= 63 g mol^{–1}

∴ Number of moles in 69 g of HNO_{3}

\begin{array}{l}=\frac{69}{63gmo{l}^{-1}}\\ =\text{1}.0\text{95 mol}\\ \text{Volume of 1}00\text{g of nitric acid solution}\\ =\frac{MassofSolution}{densityofsolution}\\ =\frac{100g}{1.41gm{L}^{-1}}\\ =70.92mL\equiv 70.92\times {10}^{-3}L\\ \text{Concentration of nitric acid}\\ =\frac{1.095mole}{70.92\times {10}^{-3}L}\\ =\text{15}.\text{44 mol}/\text{L}\\ \therefore \text{Concentration of nitric acid}=\text{15}.\text{44 mol}/\text{L}\end{array}

**Q.7 ****How much copper can be obtained from 100 g of copper sulphate (CuSO _{4})?**

**Ans.**

1 mole of CuSO_{4} contains 1 mole of copper.

Molar mass of CuSO_{4} = (63.5) + (32.00) + 4(16.00)

= 63.5 + 32.00 + 64.00

= 159.5 g

159.5 g of CuSO_{4} contains 63.5 g of copper.

⇒100 g of CuSO_{4} will contain of copper.

Amount of copper that can be obtained from 100 g CuSO_{4}

= 39.81 g

**Q.8 ****Determine the molecular formula of an oxide of iron in which the mass per cent of iron and oxygen are 69.9 and 30.1 respectively.**

**Ans.**

Mass percent of iron (Fe) = 69.9% (Given)

Mass percent of oxygen (O) = 30.1% (Given)

Number of moles of iron present in the oxide

= 1.25

Number of moles of oxygen present in the oxide

= 1.88

Ratio of iron to oxygen in the oxide,

= 1.25 :1.88

=\frac{1.25}{1.25}:\frac{1.88}{1.25}

= 1 : 1.5

= 2 : 3

The empirical formula of the oxide is Fe_{2 }O_{3}.

Empirical formula mass of Fe_{2 }O_{3}= [2(55.85) + 3(16.00)]g

Molar mass of Fe_{2O3} = 159.69 g

\therefore n=\frac{Molarmass}{Empricalformulamass}=\frac{159.69g}{159.7g}

= 0.999

= 1 (approx)

Molecular formula of a compound is obtained by multiplying the empirical formula with n.

Thus, the empirical formula of the given oxide is Fe_{2 }O_{3} and n is 1.

Hence, the molecular formula of the oxide is Fe_{2 }O_{3}.

**Q.9 ****Calculate the atomic mass (average) of chlorine using the following data:**

% Natural Abundance | Molar Mass | |

^{35}Cl |
75.77 | 34.9689 |

^{37}Cl |
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.10 ****In three moles of ethane (C _{2}H_{6}), 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 C_{2}H_{6} contains 2 moles of carbon atoms.

Number of moles of carbon atoms in 3 moles of C_{2}H_{6}

= 2 × 3 = 6

(ii) 1 mole of C_{2}H_{6} contains 6 moles of hydrogen atoms.

Number of moles of carbon atoms in 3 moles of C_{2}H_{6}

= 3 × 6 = 18

(iii) 1 mole of C_{2}H_{6} contains 6.023 × 10^{23} molecules of ethane.

Number of molecules in 3 moles of C_{2}H_{6}

= 3 × 6.023 × 10^{23} = 18.069 × 10^{23}

**Q.11 ****What is the concentration of sugar (C _{12}H_{22}O_{11}) 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\times 12\right)+\left(1\times 22\right)+\left(11\times 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.12 ****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 (CH_{3}OH) = (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,

M_{1}V_{1} = M_{2}V_{2}

(Given solution) (Solution to be prepared)

(24.78 mol L^{-1}) V_{1 }= (2.5 L) (0.25 mol L^{-1})

V_{1 }= 0.0252 L

V_{1} = 25.22 mL

**Q.13 ****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\times 9.8m{s}^{-2}}{c{m}^{2}}\times \frac{1kg}{1000g}\times \frac{{(100)}^{2}c{m}^{2}}{1{m}^{2}}

= 1.01332 × 10^{5} kg m^{–1}s^{–2}

We know,

1 N = 1 kg ms^{–2}

Then,

1 Pa = 1 Nm^{–2} = 1 kg m^{–2}s^{–2}

1 Pa = 1 kg m^{–2}s^{–2}

Pressure = 1.01332 × 10^{5}Pa

**Q.14 ****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.15 ****Match the following prefixes with their multiples:**

Prefixes | Multiples | |

(i) | Micro | 10^{6} |

(ii) | Deca | 10^{9} |

(iii) | Mega | 10^{-6} |

(iv) | Giga | 10^{-15} |

(v) | Femto | 10 |

**Ans.**

Prefixes | Multiples | |

(i) | Micro | 10^{-6} |

(ii) | Deca | 10 |

(iii) | Mega | 10^{6} |

(iv) | Giga | 10^{9} |

(v) | Femto | 10^{-15} |

**Q.16 ****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.17 ****A sample of drinking water was found to be severely contaminated with chloroform, CHCl _{3}, 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 (10^{6}) parts.

Mass percent of 15 ppm chloroform in water

=\frac{15}{{10}^{6}}\times 100

\simeq 1.5\times {10}^{-3}\%

(ii) 100 g of the sample contains 1.5 × 10^{-3} g of CHCl_{3}

⇒1000 g of the sample contains 1.5 × 10^{-2} g of CHCl_{3}

∴ Molality of chloroform in water

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

Molar mass of CHCl_{3} = 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.18 ****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 ×10^{5}

(iii) 8008 = 8.008 ×10^{3}

(iv) 500.0 = 5.000 × 10^{2}

(v) 6.0012 = 6.0012

**Q.19 ****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.20 ****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.21 ****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 = …………………. dm ^{3}**

**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\times \frac{1000m}{1km}\times \frac{100cm}{1m}\times \frac{10mm}{1cm}\\ \therefore 1km={10}^{6}mm\\ 1km=1km\times \frac{1000m}{1km}\times \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\times \frac{1g}{1000mg}\times \frac{1kg}{1000g}\\ \Rightarrow 1mg={10}^{-6}kg\\ 1mg=1mg\times \frac{1g}{1000mg}\times \frac{1kg}{{10}^{-9}g}\\ \Rightarrow 1mg={10}^{6}ng\\ \therefore 1mg={10}^{-6}kg={10}^{6}ng\\ \left(iii\right)1mL=1mL\times \frac{1L}{1000mL}\\ \Rightarrow 1mL={10}^{-3}L\\ 1mL=1c{m}^{3}=1c{m}^{3}\times \frac{1dm}{10cm}\times \frac{1dm}{10cm}\times \frac{1dm}{10cm}\\ \Rightarrow 1mL={10}^{-3}d{m}^{3}\\ \therefore 1mL={10}^{-3}L={10}^{-3}d{m}^{3}\end{array}

**Q.22 ****If the speed of light is 3.0 ×10 ^{8} 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 × 10^{8} ms^{–1}

Distance travelled by light in 2.00 ns

= Speed of light × Time taken

= (3.0 × 10^{8} ms ^{–1}) (2.00 × 10^{–9} s)

= 6.00 × 10^{–1}m

= 0.600 m

**Q.23 ****In a reaction**

**A + B _{2 }→ AB_{2}**

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

**N _{2(g)}+ 3H_{2(g) }→ 2NH_{3(g)}**

**(i) Calculate the mass of ammonia produced if 2.00 × 10 ^{3}g dinitrogen reacts with 1.00 × 10^{3} 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,

N_{2(g)}+ 3H_{2(g)}→2NH_{3(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 × 10^{3} g of dinitrogen will react with

\frac{6g}{28g}\times 2.00\times {10}^{3}g

dihydrogen i.e., 2.00 × 10^{3} g of dinitrogen will react with 428.6 g of dihydrogen.

Given,

Amount of dihydrogen = 1.00 × 10^{3} g

Hence, N_{2} is the limiting reagent.

∴ 28g of N_{2} produces 34 g of NH_{3}

Hence, mass of ammonia produced by 200 g of N_{2}

=\frac{34g}{28g}\times 2000g

= 2428.57 g

(ii) N_{2} is the limiting reagent and H_{2} is the excess reagent. Hence, H_{2} will remain unreacted.

(iii) Mass of dihydrogen left unreacted = 1.00 × 10^{3} g – 428.6 g

= 571.4 g

**Q.25 ****How are 0.50 mol Na _{2}CO_{3} and 0.50 M Na_{2}CO_{3} different?**

**Ans.**

Molar mass of Na_{2}CO_{3} = (2 × 23) + 12.00 + (3 × 16)

= 106 g mol^{–1}

Now, 1 mole of Na_{2}CO_{3} means 106 g of Na_{2}CO_{3}

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

Hence, 0.50 mol of Na_{2}CO_{3} is present in 1 L of water or 53 g of Na_{2}CO_{3} is present in 1 L of water.

**Q.26 ****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:

2H_{2(g)}+ O_{2(g) }→ 2H_{2}O_{(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.27 ****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 × 10^{4} × 10^{–3} g

Since,

1 g = 10^{–3} kg

2.5365 × 10^{1}g = 2.5365 × 10^{–1} × 10^{–3} kg

∴ 25365 mg = 2.5365 × 10^{–2} kg

**Q.28 ****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 Cl _{2}(g)**

**Ans.**

\begin{array}{l}1gofAu\left(s\right)=\frac{1}{197}molofAu(s)\\ =\frac{6.022\times {10}^{23}}{197}atomsofAu(s)\\ =\text{3}.0\text{6}\times \text{1}{0}^{\text{21}}\text{atoms of Au}\left(\text{s}\right)\\ 1gofNa\left(s\right)=\frac{1}{23}molofNa(s)\\ =\frac{6.022\times {10}^{23}}{23}atomsofNa(s)\\ =\text{}0.\text{262}\times \text{1}{0}^{\text{23}}\text{atoms of Na}\left(\text{s}\right)\\ =\text{26}.\text{2}\times \text{1}{0}^{\text{21}}\text{atoms of Na}\left(\text{s}\right)\\ 1gofLi\left(s\right)=\frac{1}{7}molofLi(s)\\ =\frac{6.022\times {10}^{23}}{7}atomsofLi(s)\\ =\text{}0.\text{86}\times \text{1}{0}^{\text{23}}\text{atoms of Li}\left(\text{s}\right)\\ =\text{86}.0\text{}\times \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}(g)\\ \left({\text{Molar mass of Cl}}_{\text{2}}\text{molecule}=\text{35}.\text{5}\times \text{2}={\text{71 g mol}}^{-\text{1}}\right)\\ =\frac{6.022\times {10}^{23}}{71}atomsofC{I}_{2}(g)\\ =\text{}0.0\text{848}\times \text{1}{0}^{\text{23}}{\text{atoms of Cl}}_{\text{2}}\left(\text{g}\right)\\ =\text{8}.\text{48}\times \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.29 ****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.30 ****What will be the mass of one ^{12}C atom in g?**

**Ans.**

1 mole of carbon atoms = 6.023 × 10^{23} atoms of carbon

= 12 g of carbon

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

= 1.993 × 10^{–23}g

**Q.31 ****How many significant figures should be present in the answer of the following calculations:**

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

**(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.32 ****Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes:**

Isotope | Isotopic molar mass | Abundance |

^{36}Ar |
35.96755 gmol^{–1} |
0.337% |

^{38}Ar |
37.96272 gmol^{–1} |
0.063% |

^{40}Ar |
39.9624 gmol^{–1} |
99.600% |

**Ans.**

Molar mass of argon

\begin{array}{l}\left[\left(35.96755\times \frac{0.337}{100}\right)+\left(37.96272\times \frac{0.063}{100}\right)+\left(39.9624\times \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.33 ****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 × 10^{23} atoms of Ar

∴ 52 mol of Ar = 52 × 6.022 × 10^{23} atoms of Ar

= 3.131 × 10^{25} 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 × 10^{23} atoms of He

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

= 7.8286 × 10^{24} atoms of He

**Q.34 ****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}\times 3.38g\\ =\text{}0.\text{9217 g}\\ \text{18 g of water contains 2 g of hydrogen}.\\ \therefore 0.690gofwaterwillcontainhydrogen=\frac{2g}{18g}\times 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}\times 100\\ =\text{92}.\text{32}\%\\ percentofHinthecompound=\frac{0.0767g}{0.9984g}\times 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}\times 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.35 ****Calcium carbonate reacts with aqueous HCl to give CaCl _{2} and CO_{2} according to the reaction, CaCO_{3}(s) + 2HCl_{(aq)} → CaCl_{2(aq)}+ CO_{2(g) }+ H_{2}O_{(l)}**

**What mass of CaCO _{3} 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}\times 25mL

= 0.6844 g

From the given chemical equation,

CaCO_{3}(s)+ 2HCl_{(aq)}→CaCl_{2(aq)}+CO_{2(g)}+H_{2}O_{(l)}

2 mol of HCl (2 × 36.5 = 73 g) react with 1 mol of CaCO_{3} (100 g).

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

=0.938 g

**Q.36 ****Chlorine is prepared in the laboratory by treating manganese dioxide (MnO _{2}) with aqueous hydrochloric acid according to the reaction**

**4HCl _{(aq) }+ MnO_{2(s) }→ 2H_{2}O_{(l)} + MnCl_{2(aq) }+ Cl_{2(g)}**

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

**Ans.**

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

[4 × 36.5 = 146 g] of HCl.

5.0 g of MnO_{2} will react with

=\frac{146g}{87g}\times 5.0gofHCL

= 8.4 g of HCl

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

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## 1. Does Extramarks provide answers to NCERT textbook questions for Class 11 Chemistry Chapter 1 in a detailed way?

Absolutely, Extramarks provides step-by-step answers to NCERT Solutions for Class 11 Chemistry Chapter 1. This allows students to study all the concepts in detail and clarify their doubts way ahead of exams.

## 2. Why is Chapter 1 of Class 11 Chemistry important?

The chapter covers basic chemistry ideas that are helpful for students in the medical and non-medical streams. Atomic mass, weight, molecular mass, temperature, and other concepts are introduced. Chemical equations and formulae are also discussed and introduced in this chapter. Students must be comprehensive with these topics because they serve as the foundation for concepts that will be studied later in the class and in Class 12. These are also crucial for board exams and other competitive exams.

## 3. Define the law of multiple proportions. Explain it with an example. How does this law prove the existence of atoms?

The Law of multiple proportions says that when two elements mix to generate two or more compounds, the masses of one of the elements which combine with a fixed mass of the other have an easy ratio to each other. e.g. Carbon and Oxygen combine to form two different chemical compounds- carbon dioxide and carbon monoxide. The masses of oxygen combined with a definite mass of carbon in CO and CO2 are 16 and 32, respectively. These masses of oxygen bear a ratio of 32: 16 or 2: 1 with one another. For example, sulphur and oxygen combine to form two compounds, namely, sulphur trioxide and sulphur dioxide.