# NCERT Solutions For Class 11 Chemistry Chapter 5

## NCERT Solutions Class 11 Chemistry States of Matter: Gases and Liquids

Class 11 is an important phase of a student's life. The topics which are taught in class 11 are the basics of the topic which will be taught in class 12. Students who wants to go for further studies after class 12 should try to understand the topics of class 11 because in competitive exams the questions are always asked from both the classes. So it is recommended to solve the NCERT questions given at the end of each chapter. NCERT Solutions for class 11 Chemistry chapter 5 states of matter is provided here for better understanding and clarification.

Q1. Calculate the minimum pressure required to compress 500 dm3$dm^{ 3 }$of air at 1 bar to 200 dm3$dm^{ 3 }$ at 30C$30^{\circ} C$ ?

Initial pressure, P1 = 1 bar

Initial volume, V1 = 500 dm3$dm^{ 3 }$

Final volume, V2 = 200 dm3$dm^{ 3 }$

As the temperature remains same,the final pressure (P2) can be calculated with the help of Boyle’s law.

Acc. Boyle’s law,

P1V1 = P2V2

P2 = P1V1V2$\frac{P_{1}V_{1}}{V_{2}}$

= 1×500200$\frac{1 \; \times \; 500}{200}$

= 2.5 bar

∴ the minimum pressure required to compress is  2.5 bar.

Q2. A container with a capacity of 120 mL contains some amount of gas at 35$35^{\circ}$ C and 1.2 bar pressure. The gas is transferred to another container of volume 180 mL at 35C$35^{\circ} C$. Calculate what will be the pressure of the gas?

Initial pressure, P1 = 1.2 bar

Initial volume, V1 = 120 mL

Final volume, V2 = 180 mL

As the temperature remains same, final pressure (P2) can be calculated with the help of Boyle’s law.

According to the Boyle’s law,

P1V1 = P2V2

P2 = P1V1V2$\frac{P_{1}V_{1}}{V_{2}}$

= 1.2×120180$\frac{1.2 \; \times \; 120}{180}$

= 0.8 bar

Therefore, the min pressure required is 0.8 bar.

Q3. Prove that at a given temp density of a gas is proportional to the gas pressure by using the equation of state pV = nRT.

The equation of state is given by,

pV = nRT ……..(1)

Where, p = pressure

V = volume

N = number of moles

R = Gas constant

T = temp

nV$\frac{n}{V}$ = pRT$\frac{p}{RT}$

Replace n with mM$\frac{m}{M}$, therefore,

mMV$\frac{m}{MV}$ = pRT$\frac{p}{RT}$……..(2)

Where, m = mass

M = molar mass

But, mV$\frac{m}{V}$ = d

Where, d = density

Therefore, from equation (2), we get

dM$\frac{d}{M}$ = pRT$\frac{p}{RT}$

d = (MRT$\frac{M}{RT}$) p

d $\propto$ p

Therefore, at a given temp, the density of gas (d) is proportional to its pressure (p).

Q4. At 0$0^{\circ}$ C, the density of a certain oxide of a gas at 2 bars is equal to that of dinitrogen at 5 bars. Calculate the molecular mass of the oxide.

Density (d) of the substance at temp (T) can be given by,

d = MpRT$\frac{Mp}{RT}$

Now, density of oxide (d1) is as given,

d1$d_{1}$ = M1p1RT$\frac{M_{1}p_{1}}{RT}$

Where, M1 = mass of the oxide

p1 = pressure of the oxide

Density of dinitrogen gas (d2) is as given,

d2$d_{2}$ = M1p2RT$\frac{M_{1}p_{2}}{RT}$

Where, M2 = mass of the oxide

p2 = pressure of the oxide

Acc to the question,

d1 = d2

Therefore, M1p1=M2p2$M_{1}p_{1} = M_{2}p_{2}$

Given:

p1$p_{1}$ = 2 bar

p2$p_{2}$ = 5 bar

Molecular mass of nitrogen, M2$M_{2}$ = 28 g/mol

Now, M1$M_{1}$

= M2p2p1$\frac{M_{ 2 }p_{2}}{p_{ 1 }}$

= 28×52$\frac{ 28 × 5 }{ 2 }$

= 70 g/mol

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

Q5. A pressure of 1 g of an ideal gas X at 27$27^{\circ}$ C is found to be 2 bars. When 2 g of another ideal gas is added in the same container at same temp the pressure becomes 3 bars. Find the relation between their molecular masses.

For ideal gas A, the ideal gas equation is given by,

pXV=nXRT$p_{X}V = n_{X}RT$……(1)

Where pX$p_{X}$ and nX$n_{X}$ represent the pressure and number of moles of gas X.

For ideal gas Y, the ideal gas equation is given by,

pYV=nYRT$p_{Y}V = n_{Y}RT$……(2)

Where, pY$p_{Y}$ and nY$n_{Y}$ represent the pressure and number of moles of gas Y.

[V and T are constants for gases X and Y]

From equation (1),

pXV=mXMX$p_{ X }V = \frac{m_{ X }}{M_{ X }}$ RT

pXMXmX$\frac{p_{ X }M_{ X }}{m_{ X }}$ = RTV$\frac{ R T}{ V }$ ……(3)

From equation (2),

pYV=mYMY$p_{ Y }V =\frac{m_{ Y }}{M_{ Y }}$ RT

pYMYmY$\frac{p_{ Y }M_{ Y }}{m_{ Y }}$ = RTV$\frac{ R T}{V}$ …… (4)

Where, MX$M_{ X }$ and MY$M_{ Y }$ are the molecular masses of gases X and Y respectively.

Now, from equation (3) and (4),

pXMXmX$\frac{p_{ X }M_{ X }}{m_{ X }}$ = pYMYmY$\frac{p_{ Y }M_{ Y }}{m_{ Y }}$ ….. (5)

Given,

mX$m_{ X }$ = 1 g

pX$p_{ X }$ = 2 bar

mY$m_{ Y }$ = 2 g

pY$p_{ Y }$ = (3 – 2) = 1 bar (Since total pressure is 3 bar)

Substituting these values in equation (5),

2×MX1$\frac{2 \; \times \; M_{X} }{1}$ = 1×MY2$\frac{1 \; \times \; M_{Y} }{2}$

4 MX$M_{ X }$ = MY$M_{ Y }$

Therefore, the relationship between the molecular masses of X and Y is,

4 MX$M_{ X }$ = MY$M_{ Y }$

Q6. The drain cleaner has small bits of aluminum, which react with caustic soda to produce dihydrogen. What volume of dihydrogen at 20$20^{\circ}$ C and 1 bar will be released when 0.15 g of aluminum reacts?

The reaction of aluminum with caustic soda is as given below:

2Al + 2NaOH + 2H2O $\rightarrow$ 2NaAlO2 + 3H2

At Standard Temperature Pressure (273.15 K and 1 atm), 54 g ( 2 × 27 g) of Al gives 3 ×22400 mL of H2.

Therefore, 0.15 g Al gives:

= 3×22400×0.1554$\frac{3 \; \times \; 22400 \; \times \; 0.15}{54}$ mL of H2

= 186.67 mL of H2

At Standard Temperature Pressure,

p1$p_{ 1 }$ = 1 atm

V1$V_{ 1 }$ = 186.67 mL

T1$T_{ 1 }$ = 273.15 K

Let the volume of dihydrogen be V2$V_{ 2 }$ at p2$p_{ 2 }$ = 0.987 atm (since 1 bar = 0.987 atm) and T2$T_{ 2 }$ = 20$20^{\circ}$ C = (273.15 + 20) K = 293.15 K.

Now,

p1V1T1$\frac{p_{ 1 }V_{ 1 }}{T_{ 1 }}$ = p2V2T2$\frac{p_{ 2 }V_{ 2 }}{T_{ 2 }}$

V2=p1V1T2p2T1$V_{ 2 } = \frac{p_{ 1 }V_{ 1 }T_{ 2 }}{p_{ 2 }T_{ 1 }}$

= 1×186.67×293.150.987×273.15$\frac{1 \; \times \; 186.67 \; \times \; 293.15}{0.987 \; \times \; 273.15}$

= 202.98 mL

= 203 mL

Hence, 203 mL of dihydrogen will be released.

Q7. Calculate the pressure exerted by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a 9 dm3$dm^{ 3 }$ at flask at 27°.

It is known that,

p = mM$\frac{m}{M}$ RTV$\frac{RT}{V}$

For methane (CH4),

pCH4$p_{CH_{ 4 }}$

= 3.216$\frac{ 3.2 }{ 16 }$ × 8.314×3009×103$\frac{8.314 \; \times \;300 }{9 \; \times \; 10^{ – 3 }}$ [Since 9 dm3 = 9×103$9 \; \times \; 10^{ -3 }$ m3         27$27^{\circ}$C = 300 K]

= 5.543 × 104$10^{ 4 }$ Pa

For carbon dioxide (CO2),

pCO2$p_{CO_{ 2 }}$

= 4.444$\frac{ 4.4 }{ 44 }$ × 8.314×3009×103$\frac{8.314 \; \times \;300 }{9 \; \times \; 10^{ – 3 }}$

= 2.771 × 104$10^{4}$ Pa

Total pressure exerted by the mixture can be calculated as:

p = pCH4$p_{CH_{ 4 }}$ + pCO2$p_{CO_{ 2 }}$

= (5.543 × 104$10^{ 4 }$ + 2.771 × 104$10^{4}$) Pa

= 8.314 × 104$10^{ 4 }$ Pa

Q8. Calculate the pressure of the gaseous mixture when 0.5 L of H2 at 0.8 bars and 2.0 L of dioxygen at 0.7 bars are introduced in a 1L container at 27$27^{\circ}$.

Let the partial pressure of H2${H_{ 2 }}$ in the container be pH2$p_{H_{ 2 }}$.

Now,

p1${p_{ 1 }}$ = 0.8 bar

p2${p_{ 2 }}$ = pH2$p_{H_{ 2 }}$

V1${V_{ 1 }}$ = 0.5 L

V2${V_{ 2 }}$ = 1 L

It is known that,

p1${p_{ 1 }}$ V1${V_{ 1 }}$ = p2${p_{ 2 }}$ V2${V_{ 2 }}$

p2${p_{ 2 }}$ = p1×V1V2$\frac{p_{ 1 } \; \times \; V_{ 1 }}{V_{ 2 }}$

pH2$p_{H_{ 2 }}$ = 0.8×0.51$\frac{ 0.8 \; \times \; 0.5 }{ 1 }$

= 0.4 bar

Now, let the partial pressure of O2 in the container be pO2$p_{O_{ 2 }}$.

Now,

p1${p_{ 1 }}$ = 0.7 bar

p2${p_{ 2 }}$ = pO2$p_{O_{ 2 }}$

V1${V_{ 1 }}$ = 2.0 L

V2${V_{ 2 }}$ = 1 L

p1${p_{ 1 }}$ V1${V_{ 1 }}$ = p2${p_{ 2 }}$ V2${V_{ 2 }}$

p2${p_{ 2 }}$ = p1×V1V2$\frac{p_{ 1 } \; \times \; V_{ 1 }}{V_{ 2 }}$

pO2$p_{O_{ 2 }}$ = 0.7×201$\frac{ 0.7 \; \times \; 20 }{ 1 }$

= 1.4 bar

Total pressure of the gas mixture in the container can be obtained as:

ptotal$p_{total}$ = pH2$p_{H_{ 2 }}$ + pO2$p_{O_{ 2 }}$

= 0.4 + 1.4

= 1.8 bar

Q9. A density of a gas is  5.46 g/dm3 at 27$27^{\circ}$C at 2 bar pressure. Calculate its density at Standard Temperature Pressure.

Given,

d1 = 5.46 g/dm3

p1 = 2 bar

T1 = 27$27^{\circ}$C = (27 + 273) K = 300 K

p2 = 1 bar

T2 = 273 K

d2 = ?

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

d = MpRT$\frac{Mp}{RT}$

d1d2$\frac{d_{1}}{d_{2}}$ = Mp1RT1Mp2RT2$\frac{\frac{M \;p_{ 1 }}{R \; T_{ 1 }}}{\frac{M \; p_{ 2 }}{R \; T_{ 2 }}}$

d1d2$\frac{d_{1}}{d_{2}}$ = p1T2p2T1$\frac{p_{1} \; T_{2}}{p_{2} \; T_{1}}$

d2 = p2T1d1p1T2$\frac{p_{2} \; T_{1} \; d_{1}}{p_{1} \; T_{2}}$

= 1×300×5.462×273$\frac{1 \; \times 300 \; \times 5.46}{2 \; \times 273}$

= 3 g dm-3

Hence, the density of the gas at STP will be 3 g dm-3

Q10. 34.05 mL of phosphorus vapour has weight 0.0625 g at 546C$546^{\circ} C$ and 0.1 bar pressure. Calculate the molar mass of phosphorus.

Given,

p = 0.1 bar

V = 34.05 mL = 34.05 × 10-3 dm3$dm^{ 3 }$

R = 0.083 bar dm3$dm^{ 3 }$ at  K-1 mol-1

T = 546C$546^{\circ} C$ = (546 + 273) K = 819 K
The no of moles (n) can be calculated using the ideal gas equation as:

pV = nRT

n = pVRT$\frac{ pV }{ RT }$

= 0.1×34.05×1030.083×819$\frac{ 0.1 \times 34.05 × 10^{-3} }{ 0.083 \times 819 }$

= 5.01 × 10-5 mol

Therefore, molar mass of phosphorus = 0.06255.01×105$\frac{ 0.0625 }{ 5.01 \times 10^{-5} }$

= 1247.5 g mol-1

Q11. A student forgot to add the reaction mixture to the container at 27$27^{\circ}$ C but instead, he placed the container on the flame. After a lapse of time, he came to know about his mistake, and using a pyrometer he found the temp of the container 477$477^{\circ}$ C. What fraction of air would have been expelled out?

Let the volume of the container be V.

The volume of the air inside the container at 27$27^{\circ}$ C is V.

Now,

V1 = V

T1 = 27$27^{\circ}$ C = 300 K V2 = ?

T2 = 477$477^{\circ}$ C = 750 K

Acc to Charles’s law,

V1T1$\frac{V_{1}}{T_{1}}$ = V2T2$\frac{V_{2}}{T_{2}}$

V1$V_{ 1 }$ = V1T2T1$\frac{V_{ 1 }T_{ 2 }}{T_{ 1 }}$

= 750V300$\frac{750V }{300 }$

= 2.5 V

Therefore, volume of air expelled out

= 2.5 V – V = 1.5 V

Hence, fraction of air expelled out

= 1.5V2.5V$\frac{1.5V }{ 2.5V }$

= 35$\frac{3 }{ 5 }$

Q12. What is the temp of 4.0 mol of gas occupying 5 dm3 at 3.32 bar?                            ( R = 0.083 bar dm3$dm^{ 3 }$ at K-1 mol-1).

Given,

N= 4.0 mol

V = 5 dm3$dm^{ 3 }$

p = 3.32 bar

R = 0.083 bar dm3$dm^{ 3 }$ at  K-1 mol-1

The temp (T) can be calculated using the ideal gas equation as:

pV = nRT

T = pVnR$\frac{ pV }{ nR }$

= 3.32×54×0.083$\frac{3.32 \; \times \; 5 }{ 4 \;\times \; 0.083 }$

= 50 K

Therefore, the required temp is 50 K.

Q13. What is the total no of electrons present in 1.4 g of dinitrogen gas?

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

Thus, 1.4 g of N2

= 1.428$\frac{ 1.4 }{ 28 }$

= 0.05 mol

= 0.05 × 6.02 × 1023 no of molecules

= 3.01 × 1023 no. of molecules

Now, 1 molecule of N2 has 14 electrons.

Therefore, 3.01 × 1023 molecules of N2 contains,

= 14 × 3.01 × 1023

= 4.214 × 1023 electrons

Q14. How much time would it take to distribute 1 Avogadro no. of wheat grains, if 1010 grains are distracted each second?

Avogadro no. = 6.02 × 1023

Therefore, time taken

= 6.02×10231010s$\frac{6.02 \; \times \; 10^{23}}{10^{10}} s$

= 6.02 × 1013 s

= 6.02×102360×60×24×365years$\frac{6.02 \; \times \; 10^{23}}{60 \; \times \; 60 \; \times \; 24 \; \times \; 365 } years$

= 1.909 × 106 years

Therefore, the time taken would be 1.909 × 106 years.

Q15. What is the total pressure in the mixture of 4 g of dihydrogen and 8 g of dioxygen in a container of 1 dm3$dm^{ 3 }$ at K-1 mol-1?

Given:

Mass of O2 = 8 g

No. of moles

= 832$\frac{ 8 }{ 32 }$

= 0.25 mole

Mass of H2 = 4 g

No. of moles

= 42$\frac{ 4 }{ 2 }$

= 2 mole

Hence, total no of moles in the mixture

= 0.25 + 2

= 2.25 mole

Given:

V = 1 dm3$dm^{ 3 }$

n = 2.25 mol

R = 0.083 bar dm3$dm^{ 3 }$ at K-1 mol-1

T = 27$27^{\circ}$ C = 300 K

Total pressure :
pV = nRT

p = nRTV$\frac{ nRT }{ V }$

= 225×0.083×3001$\frac{225 \; \times \; 0.083 \; \times \; 300 }{ 1 }$

= 56.025 bar

Therefore, the total pressure of the mixture is 56.025 bar.

Q16. The difference between the mass of displaced air and the mass of the balloon is known as pay load. What is the pay load when a balloon of radius is 10 m, mass 100 kg is filled with helium at 1.66 bar at 27$27^{\circ}$ C.

(Density of air = 1.2 kg m-3 and R = 0.083 bar dm3$dm^{ 3 }$ at  K-1 mol-1)

Given:

r = 10 m

Therefore, volume of the balloon

= 43$\frac{4}{3}$ πr3

= 43×227×103$\frac{ 4 }{ 3 }\; \times \; \frac{ 22 }{ 7 } \; \times \; 10^{3}$

= 4190.5 m3 (approx.)

Therefore, the volume of the displaced air

= 4190.5 × 1.2 kg

= 5028.6 kg

Mass of helium,

= MpVRT$\frac{ MpV }{ RT }$

Where, M = 4 × 10-3 kg mol-1

p = 1.66 bar

V = volume of the balloon

= 4190.5 m3

R = 0.083 0.083 bar dm3$dm^{ 3 }$ at K-1 mol-1

T = 27 °C = 300 K

Then,

m = 4×103×1.66×4190.5×1030.083×300$\frac{ 4 \; \times \; 10^{-3} \; \times \; 1.66 \; \times \; 4190.5 \; \times \; 10^{3}}{0.083 \; \times \; 300}$

= 1117.5 kg (approx.)

Now, total mass with helium,

= (100 + 1117.5) kg

= 1217.5 kg

= (5028.6 – 1217.5)

= 3811.1 kg

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

Q17. What is the volume occupied by 8.8 g of CO2 at 31.1$31.1^{\circ}$ C and 1 bar pressure? Given that R = 0.083 bar dm3$dm^{ 3 }$ at K-1 mol-1.

pVM = mRT

V = mRTMp$\frac{mRT}{Mp}$

Given:

m = 8.8 g

R = 0.083 bar dm3$dm^{ 3 }$ at K-1 mol-1.

T = 31.1 °C = 304.1 K

M = 44 g

p = 1 bar

Thus, Volume (V),

= 8.8×0.083×304.144×1$\frac{8.8 \; \times \; 0.083 \; \times \; 304.1}{ 44 \; \times \; 1}$

= 5.04806 L

= 5.05 L

Therefore, the volume occupied is 5.05 L.

Q18. 2.9 g of a gas at 95$95^{\circ}$ C occupied the same volume as 0.184 g of dihydrogen at 17$17^{\circ}$ C, at the same pressure. Calculate the molar mass of the gas.

Volume,

V = mRTMp$\frac{mRT}{Mp}$

= 0.184×R×2902×p$\frac{0.184 \; \times \; R \; \times \; 290}{ 2 \; \times \; p}$

Let M be the molar mass of the unknown gas.

Volume occupied by the unknown gas is,

= mRTMp$\frac{mRT}{Mp}$

= 2.9×R×368M×p$\frac{2.9 \; \times \; R \; \times \; 368}{ M \; \times \; p}$

According to the ques,

0.184×R×2902×p$\frac{0.184 \; \times \; R \; \times \; 290}{ 2 \; \times \; p}$ = 2.9×R×368M×p$\frac{2.9 \; \times \; R \; \times \; 368}{ M \; \times \; p}$

0.184×2902$\frac{0.184 \; \times \; 290}{ 2 }$ = 2.9×368M$\frac{2.9 \; \times \; 368}{ M }$

M = 2.9×368×20.184×290$\frac{2.9 \;\times \; 368 \; \times \; 2}{0.184 \; \times \; 290}$

= 40 g mol-1

Therefore, the molar mass of the gas is 40 g mol-1

Q19.  A mixture of dioxygen and dihydrogen at 1 bar pressure has 20% by weight of dihydrogen. What is the partial pressure of dihydrogen?

Let the weight of dihydrogen be 20 g.

Let the weight of dioxygen be 80 g.

No. of moles of dihydrogen (nH2),

= 202$\frac{20}{2}$

= 10 moles

No. of moles of dioxygen (nO2),

= 8032$\frac{80}{32}$

= 2.5 moles

Given:

ptotal = 1 bar

Therefore, partial pressure of dihydrogen (pH2),

= nH2nH2+nO2$\frac{ n_{H_{2}} }{ n_{H_{2}} \; + \; n_{O_{2}} }$ × ptotal

= 1010+2.5×1$\frac{ 10 }{ 10 \; + \; 2.5 } \; \times \; 1$

= 0.8 bar

Therefore, the partial pressure of dihydrogen is 0.8 bar.

Q20. What will be the SI unit for the quantity pV2T2n$\frac{ pV^{ 2 }T^{ 2 } }{ n }$?

SI unit of pressure, p = Nm2$Nm^{ -2 }$

SI unit of volume, V = m3$m^{ 3 }$

SI unit of temp, T = K

SI unit of number of moles, n = mol

Hence, SI unit of pV2T2n$\frac{ pV^{ 2 }T^{ 2 } }{ n }$ is,

= (Nm2)(m3)2(K)2mol$\frac{(Nm^{ -2 }) \; (m^{3})^{2} \; (K)^{2}}{mol}$

= Nm4K2mol1$Nm^{ 4 }K^{ 2 }mol^{ -1 }$

Q21. According to Charles’ law explain why 273$-273^{\circ}$ C is the lowest possible temp.

According to Charles’ law

At constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temp.

It was found that for all gasses (at any given pressure), the plot of volume vs. temp. (in $^{\circ}$C) is a straight line.

If we extend the line to zero volume, then it intersects the temp-axis at 273$-273^{\circ}$ C. That is the volume of any gas at273$-273^{\circ}$ C is 0. This happens because all gasses get transferred into liquid form before reaching 273$-273^{\circ}$ C.

Therefore, it can be said that273$-273^{\circ}$ C is the lowest possible temp.

Q22. Critical temp of methane and carbon dioxide are 81.9$-81.9^{\circ}$ C and 31.1C$31.1^{\circ} C$ respectively. Which of the following have stronger intermolecular forces? Why?

If the critical temp of a gas is higher then it is easier to liquefy. That is the intermolecular forces of attraction among the molecules of gas are directly proportional to its critical temp.

Therefore, in CO2 intermolecular forces of attraction are stronger.

Q23. What is the physical significance of Van der Waals parameters?