NCERT Solutions For Class 11 Chemistry Chapter 5

Q.1 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,p_2=\frac{p_1{v}_1}{v_2}\\ =\frac{1\times 500}{200}bar\\ =2.5\text{bar}\end{array}

Therefore, the minimum pressure required is 2.5 bar.

Q.2 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\times 120}{180}bar\\ =0.8\text{bar}\end{array}

Therefore, the pressure would be 0.8 bar.

Q.3 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}\mathrm{\dots \dots \dots \dots }(ii)\\ \mathrm{Re}placingn\text{with}\frac{m}{M}\text{in equation (ii), we have}\\ \frac{m}{MV}=\frac{p}{RT}\mathrm{\dots \dots \dots \dots }(iii)\end{array}

Where,

m →Mass of gas

M →Molar mass of gas

\begin{array}{l}But\frac{m}{MV}=d\left(d=density\text{of gas}\right)\\ \text{from equation}\left(\text{iii}\right)\\ \frac{d}{M}=\frac{p}{RT}\\ \Rightarrow 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\tan t\\ d=\left(cons\tan t\right)p\\ \Rightarrow d\propto p\end{array}

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

Q.4 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\times 5}{2}\end{array}

=70 g/mol

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

Q.5 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 = n_ART

Here, p_A and n_A 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\mathrm{\dots \dots \dots \dots .}(ii)

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\\ \Rightarrow \frac{p_A{M}_A}{m_A}=\frac{RT}{V}\mathrm{\dots \dots \dots }(iii)\end{array}

From equation (ii),

\begin{array}{l}{p}_{B}V=\frac{m_B}{M_B}RT\\ \Rightarrow \frac{p_B{M}_B}{m_B}=\frac{RT}{V}\mathrm{\dots \dots \dots }(iv)\end{array}

Here, M_A and M_B 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\\ \Rightarrow \frac{p_B{M}_B}{m_B}=\frac{RT}{V}\mathrm{\dots \dots \dots }(iv)\\ 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\times M_A}{1}=\frac{1\times M_B}{2}\\ \Rightarrow 4M_A=M_B\\ \text{Thus},\text{the relationship of molecular masses of A and B can be given as}:\\ 4M_A=M_B\end{array}

Q.6 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\times 27g\end{array}}{2Al}+2NaOH+2H_{2}O\to 2NaAI{O}_2+\underset{\begin{array}{l}\\ 3\times 22400mL\end{array}}{3H_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\times 22400\times 0.15}{54}mL{\text{of H}}_{2}i.e.,186.67{\text{mL of H}}_2

At STP,

\begin{array}{l}{p}_1=1atm\\ V_1=186.67\text{mL}\\ {\text{T}}_1=273.15K\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}\\ \Rightarrow V_2=\frac{p_1{V}_1{T}_2}{p_2{T}_1}\\ =\frac{1\times 186.67\times 293.15}{0.987\times 273.15}\\ =202.98\text{mL}\\ \text{=203 mL}\end{array}

Therefore, 203 mL of dihydrogen will be released.

Q.7 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}\times \frac{8.314\times 300}{9\times {10}^{-3}}\left[\frac{Since9d{m}^3=9\times {10}^{-3}{m}^3}{27ºC=300K}\right]\\ =5.543\times {10}^{4}pa\\ \text{For carbon dioxide}\left({\text{CO}}_{\text{2}}\right),\\ p_{C{O}_2}=\frac{4.4}{44}\times \frac{8.314\times 300}{9\times {10}^{-3}}\\ =2.771\times {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\times {10}^4+2.771\times {10}^4\right)pa\\ =8.314\times {10}^{4}pa\end{array}

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

Q.8 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 P_H2

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\Rightarrow 0.8\times 0.5=P_2\times 1\\ \Rightarrow \therefore P_2=p_{H_2}=0.4\text{bar}\\ \text{Similarly}\mathrm{\dots }\text{Partial pressure of dioxygen in 1 L container}\\ P_1{V}_1=P_2{V}_2\Rightarrow 0.7\times 2.0=P_2\times 1\\ \Rightarrow \therefore P_2=p_{O_2}=1.4\text{bar}\\ \text{Therefore}\mathrm{\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.9 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 \frac{d_1}{d_2}=\frac{\frac{M{p}_1}{R{T}_1}}{\frac{M{p}_2}{R{T}_2}}\\ \Rightarrow \frac{d_1}{d_2}=\frac{p_1{T}_2}{p_2{T}_1}\\ \Rightarrow d_2=\frac{p_2{T}_1{d}_1}{p_1{T}_2}\\ =\frac{1\times 300\times 5.46}{2\times 273}\\ =3\text{g}d{m}^{-3}\end{array}

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

Q.10 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\\ \Rightarrow n=\frac{pV}{RT}\\ =\frac{0.1\times 34.05\times {10}^{-3}}{0.083\times 819}\\ =5.01\times {10}^{-5}\text{mol}\\ \text{Molar mass of phosphorus}=\frac{0.0625}{5.01\times {10}^{-5}}=124.75{\text{g mol}}^{-1}\end{array}

The molar mass of phosphorus is 124.75 g mol^–1

Q.11 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}\\ \Rightarrow 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.12 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}\Rightarrow T=\frac{pV}{nR}\\ =\frac{3.32\times 5}{4\times 0.083}\\ =50\text{K}\end{array}

Temperature is 50 K.

Q.13 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.14 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\times {10}^{23}}{{10}^{10}}S\\ =6.02\times {10}^{23}S\\ =\frac{6.02\times {10}^{23}}{60\times 60\times 24\times 365}years\\ =1.909\times {10}^{6}years\\ Time\text{taken}=1.909\times {10}^{6}years\end{array}

Q.15 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.wt}\\ O_2=\frac{8}{32}=0.25\text{mole}\\ \text{Mass of dihydrogen}\left({\text{H}}_{\text{2}}\right)=\text{4 g (Given)}\\ No.{\text{of moles of H}}_2=\frac{Mass}{M.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,p=\frac{nRT}{V}\\ =2.25\times 0.083\times 3001\\ =56.025\text{bar}\end{array}

The total pressure of the mixture is 56.025 bar.

Q.16 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{Volume of ballon}=\frac{4}{3}\pi r^3\\ =\frac{4}{3}\times \frac{22}{7}\times {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\times {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=300K\\ m=\frac{4\times {10}^{-3}\times 1.66\times 4190.5\times {10}^3}{0.083\times 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.17 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\\ \Rightarrow 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\times 0.083\times 304.1}{44\times 1}

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

Q.18 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}\times \frac{R\times 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\times 368}{p}\\ \text{According to the question,}\\ \frac{0.184}{2}\times \frac{R\times 290}{p}=\frac{2.9}{M}\frac{R\times 368}{M}\\ \Rightarrow \frac{0.184\times 290}{2}=\frac{2.9\times 368}{M}\\ \Rightarrow M=\frac{2.9\times 368\times 2}{0.184\times 290}\\ =40g{\text{mol}}^{-1}\end{array}

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

Q.19 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, p_total= 1 bar
Then, partial pressure of dihydrogen,

\begin{array}{l}{p}_{H_2}=\frac{n_{H_2}}{n_{H_2}+n_{O_2}}\times p_{total}\\ =\frac{10}{10+2.5}\times 1\\ =0.8\text{bar}\end{array}

The partial pressure of dihydrogen is 0.8 bar.

Q.20 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.21 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.22 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.23 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.