# NCERT Solutions For Class 11 Chemistry chapter 5-States Of Matter

**NCERT Solutions for class-11 Chemistry** Chapter 5 States of Matter is prepared by our senior and renown teachers of Physics Wallah primary focus while solving these questions of class-11 in NCERT textbook, also do read theory of this Chapter 5 States of Matter while going before solving the NCERT questions. Our Physics Wallah team Prepared Other Subjects **NCERT Solutions** for class 11.

### Chapter 5 States of Matter

### Answer the following Questions.

**1. What will be the minimum pressure required to compress 500 dm ^{3} of air at 1 bar to 200 dm^{3}^{ }at 30°C?**

**Solution :**

Given,

Initial pressure, P

_{1}= 1 bar

Initial volume, V

_{1 }= 500 dm

^{3}

Final volume, V

_{2}= 200 dm

^{3}

Since the temperature remains constant, the final pressure (P

_{2}) can be calculated using Boyle's law.

According to Boyle's law,

**2. A container with a capacity of 120 mL contains some amount of gas at 35∘ C and 1.2 bar pressure. The gas is transferred to another container of volume 180 mL at 35∘C. Calculate what will be the pressure of the gas?**

**Solution :**

Initial pressure, P_{1} = 1.2 bar

Initial volume, V_{1} = 120 mL

Final volume, V_{2} = 180 mL

As the temperature remains same, final pressure (P_{2}) can be calculated with the help of Boyle’s law.

According to the Boyle’s law,

Therefore, the min pressure required is 0.8 bar.

**3. Using the equation of state pV = nRT; show that at a given temperature density of a gas is proportional to gas pressurep.**

**Solution :**

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,

Replacing n with m/M, we have

Where,

m → Mass of gas

M → Molar mass of gas

But, m/v= d (d = density of gas)

Thus, from equation (ii), we have

Molar mass (M) of a gas is always constant and therefore, at constant temperature= constant.

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

**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?**

**Solution :**

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

d =

Now, density of oxide (d_{1}) is given by,

Where, M_{1} and p_{1} are the mass and pressure of the oxide respectively.

Density of dinitrogen gas (d_{2}) is given by,

Where, M_{2} and p_{2} are the mass and pressure of the oxide respectively.

According to the given question,

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

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

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

**Solution :**

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

Where, p_{A} and n_{A }represent the pressure and number of moles of gas A.

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

Where, p_{B} and n_{B} represent the pressure and number of moles of gas B.

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

From equation (i), we have

From equation (ii), we have

Where, M_{A} and M_{B} are the molecular masses of gases A and B respectively.

Now, from equations (iii) and (iv), we have

Given,

(Since total pressure is 3 bar)

Substituting these values in equation (v), we have

Thus, a relationship between the molecular masses of A and B is given by.

**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?**

**Solution :**

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

At STP (273.15 K and 1 atm), 54 g (2 × 27 g) of Al gives 3 × 22400 mL of H_{2}..

0.15 g Al gives i.e., 186.67 mL of H_{2}.

At STP,

Let the volume of dihydrogen be at p_{2} = 0.987 atm (since 1 bar = 0.987 atm) and T_{2} = 20°C = (273.15 + 20) K = 293.15 K..

Therefore, 203 mL of dihydrogen will be released.

**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 dm3 flask at 27 °C ?**

**Solution :**

It is known that,

For methane (CH_{4}),

For carbon dioxide (CO_{2}),

Total pressure exerted by the mixture can be obtained as:

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

**8. What will be the pressure of the gaseous mixture when 0.5 L of H _{2} at 0.8 bar and 2.0 L of dioxygen at 0.7 bar are introduced in a 1L vessel at 27°C?**

**Solution :**

Let the partial pressure of H_{2 }in the vessel be.

Now,

= ?

It is known that,

Now, let the partial pressure of O_{2} in the vessel be.

p_{1}V_{1}= p_{2} V_{2}⇒p_{2} =p_{1} V_{1 }/V2

⇒pO_{2}=0.7×2 .0 /1= 1.4 bar

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

Hence, the total pressure of the gaseous mixture in the vessel is.

**9. Density of a gas is found to be 5.46 g /dm ^{3} at 27 °C at 2 bar pressure. What will be its density at STP?**

**Solution :**

Given,

The density (d_{2}) of the gas at STP can be calculated using the equation,

Hence, the density of the gas at STP will be 3 g dm^{–3}.

**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?**

**Solution :**

Given,

p = 0.1 bar

V = 34.05 mL = 34.05 × 10^{–3} L = 34.05 × 10^{–3} dm^{3}

R = 0.083 bar dm^{3} K^{–1} mol^{–1}

T = 546°C = (546 + 273) K = 819 K

The number of moles (n) can be calculated using the ideal gas equation as:

Therefore, molar mass of phosphorus = 1247.5 g mol^{–1}

Hence, the molar mass of phosphorus is 1247.5 g mol^{–1}.

**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?**

**Solution :**

Let the volume of the round bottomed flask be V.

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

Now,

V_{1} = V

T_{1 }= 27°C = 300 K

V_{2} =?

T_{2} = 477° C = 750 K

According to Charles’s law,

Therefore, volume of air expelled out = 2.5 V – V = 1.5 V

Hence, fraction of air expelled out =3/5

**12. Calculate the temperature of 4.0 mol of a gas occupying 5 dm ^{3} at 3.32 bar.(R = 0.083 bar dm3 K^{–1} mol^{–1}).**

**Solution :**

Given,

n = 4.0 mol

V = 5 dm

^{3}

p = 3.32 bar

R = 0.083 bar dm3 K

^{–1}mol

^{–1}

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

Hence, the required temperature is 50 K.

**13. Calculate the total number of electrons present in 1.4 g of dinitrogen gas.**

**Solution :**

Molar mass of dinitrogen (N

_{2}) = 28 g mol

^{–1}

Thus, 1.4 g of

Now, 1 molecule of contains 14 electrons.

Therefore, 3.01 × 10

^{23}molecules of N

_{2}contains = 1.4 × 3.01 × 1023

= 4.214 × 10

^{23}electrons

**14. How much time would it take to distribute one Avogadro number of wheat grains, if 10**

^{10}grains are distributed each second?
**Solution :**

Avogadro number = 6.02 × 10^{23}

Thus, time required

=

Hence, the time taken would be .

**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}.**

**Solution :**

Given,

Mass of dioxygen (O

_{2}) = 8 g

Thus, number of moles of

Mass of dihydrogen (H

_{2}) = 4 g

Thus, number of moles of

Therefore, total number of moles in the mixture = 0.25 + 2 = 2.25 mole

Given,

V = 1 dm

^{3}

n = 2.25 mol

R = 0.083 bar dm3 K

^{–1}mol

^{–1}

T = 27°C = 300 K

Total pressure (p) can be calculated as:

pV = nRT

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

**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^{–1} mol^{–1}).**

**Solution :**

Given,

Radius of the balloon, r = 10 m

Volume of the balloon

Thus, the volume of the displaced air is 4190.5 m

^{3}.

Given,

Density of air = 1.2 kg m

^{–3}

Then, mass of displaced air = 4190.5 × 1.2 kg

= 5028.6 kg

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

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.

**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}.**Solution :**

It is known that,

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

Hence, the volume occupied is 5.05 L.

**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?**

**Solution :**

Volume (V) occupied by dihydrogen is given by,

Let M be the molar mass of the unknown gas. Volume (V) occupied by the unknown gas can be calculated as:

According to the question,

Hence, the molar mass of the gas is 40 g mol

^{–1}.

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

**Solution :**

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

Then, the number of moles of dihydrogen, and the number of moles of dioxygen, .

Given,

Total pressure of the mixture, p

_{total}= 1 bar

Then, partial pressure of dihydrogen,

Hence, the partial pressure of dihydrogen is.

**20. What would be the SI unit for the quantity pV**

_{2}T_{ 2}/n?**Solution :**

The SI unit for pressure, p is Nm

^{–2}.

The SI unit for volume, V is m

^{3}.

The SI unit for temperature, T is K.

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

Therefore, the SI unit for quantity is given by,

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

**Solution :**

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

It was found that for all gases (at any given pressure), the plots of volume vs. temperature (in °C) is a straight line. If this line is extended to zero volume, then it intersects the temperature-axis at – 273°C. In other words, the volume of any gas at –273°C is zero. This is because all gases get liquefied before reaching a temperature of – 273°C. Hence, it can be concluded that – 273°C is the lowest possible temperature.

**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?**

**Solution :**

Higher is the critical temperature of a gas, easier is its liquefaction. This means that the intermolecular forces of attraction between the molecules of a gas are directly proportional to its critical temperature. Hence, intermolecular forces of attraction are stronger in the case of CO2.

**23. Explain the physical significance of Van der Waals parameters.**

**Solution :**

Physical significance of ‘a’:

‘a’ is a measure of the magnitude of intermolecular attractive forces within a gas.

Physical significance of ‘b’:

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

## NCERT Solutions For Class 11 Chemistry Chapter Wise.

#### Recent Concepts

- chapter 1-Some Basic Concepts of Chemistry
- Chapter 2- Structure of atom
- chapter 3-Classi fication of Elements and Periodicity in Properties
- chapter 4-Chemical Bonding And Molecular Structure
- chapter 5-States Of Matter
- chapter 6- Thermodynamics
- chapter 7-Equilibrium
- chapter 8-Redox Reactions
- chapter 9- Hydrogen
- chapter 10-S-Block elements
- chapter 11-p-Block Elements
- chapter 12-Some Basic Principles and Techniques
- chapter 13- Hydrocarbons
- chapter 14- Environmental Chemistry