States of Matter Class 11 Chapter 5 Questions And Answers

NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 exams must go through NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter. Going through the solutions provided on this page will help you to know how to approach and solve the problems.

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Class 11 Chemistry Chapter 5 States of Matter NCERT Solutions

States of Matter NCERT Solutions

NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter are been solved by expert teachers of CBSETuts.com. All the solutions given in this page are solved based on CBSE Syllabus and NCERT guidelines.

NCERT Exercises

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,

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.

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,

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.

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.

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

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.

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

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

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

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,

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.

Question 13.
Calculate the total number of electrons present in 1.4 g of dinitrogen gas.
Solution.
Number of moles of dinitrogen

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
\(=\frac { 1\times 6.022\times { 10 }^{ 23 } }{ { 10 }^{ 10 } } =6.022\times { 10 }^{ 23 }\quad sec\)
\(=\frac { 6.022\times { 10 }^{ 13 } }{ 60\times 60\times 24\times 365 } =1.90956\times { 10 }^{ 6 }\quad years\)

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,

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.

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

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

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

Question 20.
What would be the SI unit for the quantity pV²T²/n?
Solution.

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.

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.

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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