NCERT Solutions | Class 11 Chemistry Chapter 5

NCERT Solutions | Class 11 Chemistry Chapter 5 | States of Matter 

NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter

CBSE Solutions | Chemistry Class 11

Check the below NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter Pdf free download. NCERT Solutions Class 11 Chemistry  were prepared based on the latest exam pattern. We have Provided States of Matter Class 11 Chemistry NCERT Solutions to help students understand the concept very well.

NCERT | Class 11 Chemistry

NCERT Solutions Class 11 Chemistry
Book: National Council of Educational Research and Training (NCERT)
Board: Central Board of Secondary Education (CBSE)
Class: 11th
Subject: Chemistry
Chapter: 5
Chapters Name: States of Matter
Medium: English

States of Matter | Class 11 Chemistry | NCERT Books Solutions

You can refer to MCQ Questions for Class 11 Chemistry Chapter 5 States of Matter to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.

NCERT Exercises

NCERT Solutions for Class 11 Chemistry Chapter 5, Question 1.

What will be the minimum pressure required to compress 500 dm3 of air at 1 bar to 200 dm3 at 30°C?

Solution.

According to Boyle’s law, at constant temperature,
NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter 1

NCERT Solutions for Class 11 Chemistry Chapter 5, Question 2.

A vessel of 120 mL capacity contains a certain amount of gas of 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?

Solution.

Since temperature and amount of gas remain constant, therefore, Boyle’s law is applicable.
NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter 2

NCERT Solutions for Class 11 Chemistry Chapter 5, Question 3.

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

Solution.

According to ideal gas equation for ‘n’ moles of a gas,
NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter 3

NCERT Solutions for Class 11 Chemistry Chapter 5, Question 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.

NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter 4

NCERT Solutions for Class 11 Chemistry Chapter 5, Question 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.

NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter 5

NCERT Solutions for Class 11 Chemistry Chapter 5, Question 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 0.15 g of aluminium reacts?

Solution.

The reaction between aluminium and caustic soda is
NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter 6

NCERT Solutions for Class 11 Chemistry Chapter 5, Question 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.

NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter 7

NCERT Solutions for Class 11 Chemistry Chapter 5, Question 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 1 L vessel at 27°C?

Solution.

Partial pressure of hydrogen gas
NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter 8

NCERT Solutions for Class 11 Chemistry Chapter 5, Question 9.

Density of a gas is found to be 5.46 g/dm3 at 27°C and at 2 bar pressure. What will be its density at STP?

Solution.

We know that PV = nRT
NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter 9

NCERT Solutions for Class 11 Chemistry Chapter 5, Question 10.

34.05 mL of phosphorus vapours weigh 0.0625 g at 546°C and 0.1 bar pressure. What is the molar mass of phosphorus?

Solution.

Molar mass of phosphorus, M
NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter 10

NCERT Solutions for Class 11 Chemistry Chapter 5, Question 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.

Suppose the number of moles of gas present at 27°C in flask of volume V at pressure P is n1 then assuming ideal gas behaviour,
NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter 11

NCERT Solutions for Class 11 Chemistry Chapter 5, Question 12.

Calculate the temperature of 4.0 mol of a gas occupying 5 dm3 at 3.32 bar.
(R – 0.083 bar dm3 K-1 mol-1)

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter 12

NCERT Solutions for Class 11 Chemistry Chapter 5, Question 13.

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

Solution.

Number of moles of dinitrogen
NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter 13

NCERT Solutions for Class 11 Chemistry Chapter 5, Question 14.

How much time would it take to distribute one Avogadro number of wheat grains, if 1010 grains are distributed each second?

Solution.

Time taken to distribute 1010 grains = 1 sec.
Time taken to distribute 6.023 x 1023 grains
\(=\frac { 1\times 6.022\times { 10 }^{ 23 } }{ { 10 }^{ 10 } } =6.022\times { 10 }^{ 23 }\quad sec\)
\(=\frac { 6.022\times { 10 }^{ 13 } }{ 60\times 60\times 24\times 365 } =1.90956\times { 10 }^{ 6 }\quad years\)

NCERT Solutions for Class 11 Chemistry Chapter 5, Question 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.

Partial pressure of oxygen gas,
NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter 14

NCERT Solutions for Class 11 Chemistry Chapter 5, Question 16.

Payload is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the payload 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.

NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter 15

NCERT Solutions for Class 11 Chemistry Chapter 5, Question 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.

According to ideal gas equation, PV= nRT
NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter 16

NCERT Solutions for Class 11 Chemistry Chapter 5, Question 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.

Let molar mass of gas = M
NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter 17

NCERT Solutions for Class 11 Chemistry Chapter 5, Question 19.

A mixture of dihydrogen and dioxygen at one bar pressurecontains 20% by weight of dihydgrogen. Calculate the partial pressure of dihydrogen.

Solution.

Let the total mass of the mixture be 100 g
NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter 18
NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter 19

NCERT Solutions for Class 11 Chemistry Chapter 5, Question 20.

What would be the SI unit for the quantity pV2T2/n?

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter 20

NCERT Solutions for Class 11 Chemistry Chapter 5, Question 21.

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

Solution.

According to Charles’ law,
Volume of gas at 1°C,
\({ V }_{ t }={ V }_{ 0 }\left[ 1+\frac { t }{ 273 } \right] \) (∵ V0 = Volume at 0°C)
Volume of gas at -273°C, \(V={ V }_{ 0 }\left[ 1+\frac { 273 }{ 273 } \right] =0\)
Thus, -273°C is the lowest possible temperature because below this temperature, the volume will become negative, and that is meaningless. This lowest temperature is called absolute zero temperature.

NCERT Solutions for Class 11 Chemistry Chapter 5, Question 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 the critical temperature, more easily the gas can be liquefied i.e. greater are the intermolecular forces of attraction, CO2 has stronger intermolecular forces than methane.

NCERT Solutions for Class 11 Chemistry Chapter 5, Question 23.

Explain the physical significance of van der Waals parameters.

Solution.

van der Waals constant ‘a’ is a measure of the magnitude of the attractive forces among the molecules of a gas, while constant ‘b’ is a measure of effective size of the gas molecules.

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