**Class 11 Chemistry Chapter 5 States of Matter**

**NCERT Solutions For Class 11 Chemistry Chapter 5 States of Matter, (Chemistry) exam are Students are taught thru NCERT books in some of the state board and CBSE Schools. As the chapter involves an end, there is an exercise provided to assist students to prepare for evaluation. Students need to clear up those exercises very well because the questions withinside the very last asked from those.**

**Sometimes, students get stuck withinside the exercises and are not able to clear up all of the questions. To assist students to solve all of the questions and maintain their studies without a doubt, we have provided step-by-step NCERT Solutions for the students for all classes. These answers will similarly help students in scoring better marks with the assist of properly illustrated solutions as a way to similarly assist the students and answering the questions right.**

**NCERT Solutions For Class 11 Chemistry Chapter 5 States of Matter**

**Class 11 Chemistry Chapter 5 States of Matter**

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

**Answer: **

P_{1} = 1 bar,P_{2} = ? V_{1}= 500 dm^{3} ,V_{2}=200 dm^{3}

As temperature remains constant at 30°C,

P_{1}V_{1}=P_{2}V_{2}

1 bar x 500 dm^{3} = P_{2} x 200 dm^{3} or P_{2}=500/200 bar=2.5 bar

**5.2. 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?**

**Answer: **

V_{1}= 120 mL, P_{1}=1.2 bar,

V_{2} = 180 mL, P_{2} = ?

As temperature remains constant, P_{1}V_{1} = P_{2}V_{2}

(1.2 bar) (120 mL) = P2 (180mL)

**5.3. Using the equation of state PV = nRT, show that at a given temperature, density of a gas is proportional to the gas pressure P.**

**Answer:**

According to ideal gas equation

PV = nRT or PV=nRT/V

**5.4. At 0°C, the density of a gaseous oxide at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?**

**Answer:**

Using the expression, d =MP/RT , at the same temperature and for same density,

M_{1}P_{1} = M_{2}P_{2} (as R is constant)

(Gaseous oxide) (N_{2})

or

M_{1} x 2 = 28 x 5(Molecular mass of N_{2} = 28 u)

or M_{1} = 70u

**5.5. Pressure of l 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 the relationship between their molecular masses.**

**Answer:**

Suppose molecular masses of A and B are M_{A} and M_{B} respectively. Then their number of moles will be

**5.6. The drain cleaner, Drainex contains small bits of aluminium 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 aluminium reacts?**

**Answer:**

The chemical equation for the reaction is

2 Al + 2 NaOH + H_{2}0 -> 2 NaAl0_{2} + 3H_{2} (3 x 22400 mL At N.T.P)

2 x 27 = 54 g.

54 g of Al at N.T.P release

H_{2} gas = 3 x 22400 0.15 g of Al at N.T.P release

**5.7. What will be the pressure exerted by a mixture of 3.2g of methane and 4.4g of carbon dioxide contained in a 9 dm ^{3} flask at 27 °C?**

**Answer:**

**5.8. What will be the pressure of the gas 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 all vessel at 27 °C?**

**Answer: **

Calculation of partial pressure of H_{2} in 1L vessel P_{1}= 0.8 bar,

P_{2}= ? V_{1}= 0.5 L , V_{2} = 1.0 L

As temperature remains constant, P_{1}V_{1} = P_{2}V_{2}

(0.8 bar) (0.5 L) = P_{2} (1.0 L) or P_{2} = 0.40 bar, i.e., PH_{2} = 0.40 bar

Calculation of partial pressure of 02 in 1 L vessel

P_{1}‘ V_{1} = P_{2}‘V_{2}‘

(0.7 bar) (2.0 L) = P_{2} (1L) or P_{2}‘ = 1.4 bar, i.e.,Po_{2}= 1.4 bar

Total pressure =P_{Hz} + P_{Q2} = 0.4 bar + 1.4 bar = 1.8 bar

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

**Answer:**

**5.10. 34.05 mL of phosphorus vapour weighs 0.0625 g at 546°C and 1.0 bar pressure. What is the molar mass of phosphorus?**

**Answer:**

**5.11. 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?**

**Answer:**

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

**Answer:**

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

**Answer:**

Molecular mass of N_{2 }= 28g

28 g of N_{2} has No. of molecules = 6.022 x 10^{23} 1.4 g of

N_{2} has No. of molecules = 6.022 x 10^{23} x 1.4 g/28 g

= 3.011 x 10^{22} molecules.

Atomic No. of Nitrogen (N) = 7

1 molecule of N_{2} has electrons = 7 x 2 = 14

3.011 x 10^{22} molecules of N_{2} have electrons

= 14 x 3.011 x 10^{22}

= 4.215 x 10^{23} electrons.

**5.14. How much time would it take to distribute one Avogadro number of wheat grains if 10 ^{10} grains are distributed each second ?**

**Answer:**

**5.15. Calculate the total pressure in a mixture of 8g of oxygen and 4g of hydrogen confined in a vessel of l dm ^{3} at 27°C. R = 0.083 bar dm^{3} K^{-1} mol^{-1}.**

**Answer:**

**5.16. Pay load is defined as the difference between the mass of the 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 dm^{3} K^{-1} mol^{-1}).**

**Answer:**

**5.17. Calculate the volume occupied by 8.8 g of CO _{2} at 31.1 °C and 1 bar pressure. R = 0.083 bar LK^{-1} mol^{-1}**

**Answer:**

**5.18. 2.9 g of a gas at 95°C occupied the same volume as 0.184 g of hydrogen at 17°C at the same pressure. What is the molar mass of the gas ?**

**Answer:**

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

**Answer: **

As the mixture H_{2 }and O_{2} contains 20% by weight of dihydrogen, therefore, if H_{2} = 20g, then O_{2} = 80g

**5.20. What would be the SI unit for the quantity PV ^{2}T^{2}/n?**

**Answer:**

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

**Answer: **

At -273°C, volume of the gas becomes equal to zero, i.e., the gas ceases to exist.

**5.22. Critical temperature for Co _{2} and CH_{4} are 31.1°C and -81.9°C respectively. Which of these has stronger intermolecular forces and why ?**

**Answer: **

Higher the critical temperature, more easily the gas can be liquefied, i.e., greater are the intermolecular forces of attraction. Hence, Co_{2} has stronger intermolecular forces than CH_{4}.

**5.23. Explain the physical significance of vander Waals parameters.**

**Answer:**

‘a’ is a pleasure of the magnitude of the intermolecular forces of attraction, while b is a measure of the effective size of the gas molecules.