# NCERT SOLUTIONS FOR CLASS 11 CHEMISTRY CHAPTER 5-STATES OF MATTER￼

Welcome to Swastik Classes! We are delighted to present to you the comprehensive NCERT Solutions for Class 11 Chemistry Chapter 5 – States of Matter. In this chapter, we delve into the intriguing world of different states of matter and their properties.

Our meticulously crafted NCERT Solutions provide a step-by-step approach to understanding the fundamental concepts and principles of states of matter. Whether you are a student seeking clarity or an educator looking for reliable teaching resources, our solutions offer a comprehensive and easy-to-follow guide.

Through our well-explained explanations, examples, and practice questions, we aim to help you build a solid foundation in understanding the three states of matter: solids, liquids, and gases. We explore topics such as the kinetic molecular theory, intermolecular forces, gas laws, and much more.

Our team of experienced educators and subject matter experts has designed these solutions with a student-centric approach, ensuring that you not only grasp the theoretical aspects but also develop problem-solving skills. By engaging with our solutions, you will gain the confidence to tackle a wide range of questions and excel in your examinations.

At Swastik Classes, we believe in making learning enjoyable and accessible. Our NCERT Solutions are designed to simplify complex concepts, foster critical thinking, and stimulate your curiosity about the fascinating world of states of matter. We provide you with the necessary tools to excel academically and develop a lifelong love for the subject.

Embrace the enriching experience of learning about states of matter with our NCERT Solutions for Class 11 Chemistry Chapter 5. Let Swastik Classes be your trusted companion on your educational journey. Get ready to unlock the mysteries of different states of matter and achieve academic excellence!

## NCERT SOLUTIONS FOR CLASS 11 CHEMISTRY CHAPTER 5-STATES OF MATTER – Exercises

### Chapter-5 States of Matter

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

Solution :
Given,
Initial pressure, P1 = 1 bar
Initial volume, V= 500  dm3
Final volume, V2 = 200 dm3
Since the temperature remains constant, the final pressure (P2) 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, 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,

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 (d1) is given by,

Where, M1 and p1 are the mass and pressure of the oxide respectively.
Density of dinitrogen gas (d2) is given by,

Where, M2 and p2 are the mass and pressure of the oxide respectively.
According to the given question,

Molecular mass of nitrogen, M2 = 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, pA and nrepresent the pressure and number of moles of gas A.
For ideal gas B, the ideal gas equation is given by,

Where, pB and nB 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, MA and MB 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 H2..
0.15 g Al gives i.e., 186.67 mL of H2.
At STP,
Let the volume of dihydrogen be at p2 = 0.987 atm (since 1 bar = 0.987 atm) and T2 = 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 (CH­4),

For carbon dioxide (CO2),

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 H2 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 Hin the vessel be.
Now,

= ?

It is known that,

Now, let the partial pressure of O2 in the vessel be.

p1V1= p2 V2⇒p2 =p1 V/V2

⇒pO2=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 /dm3 at 27 °C at 2 bar pressure. What will be its density at STP?

Solution :
Given,

The density (d2) 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 dm3
R = 0.083 bar dm3 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,
V1 = V
T= 27°C = 300 K
V2 =?
T2 = 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 dm3 at 3.32 bar.(R = 0.083 bar dm3 K–1 mol–1).

Solution :
Given,
n = 4.0 mol
V = 5 dm3
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 (N2) = 28 g mol–1
Thus, 1.4 g of

Now, 1 molecule of contains 14 electrons.
Therefore, 3.01 × 1023 molecules of N2 contains = 1.4 × 3.01 × 1023
= 4.214 × 1023 electrons

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

Solution :
Avogadro number = 6.02 × 1023
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 (O2) = 8 g
Thus, number of moles of
Mass of dihydrogen (H2) = 4 g
Thus, number of moles of
Therefore, total number of moles in the mixture = 0.25 + 2 = 2.25 mole

Given,
V = 1 dm3
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 m3.
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,andthe number of moles of dioxygen, .
Given,
Total pressure of the mixture, ptotal = 1 bar
Then, partial pressure of dihydrogen,

Hence, the partial pressure of dihydrogen is.

20. What would be the SI unit for the quantity pV2T 2/n?

Solution :
The SI unit for pressure, p is Nm–2.
The SI unit for volume, V is m3.
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.

## Conclusions for NCERT SOLUTIONS FOR CLASS 11 CHEMISTRY CHAPTER 5-STATES OF MATTER

The NCERT Solutions for Class 11 Chemistry Chapter 5 – States of Matter offered by Swastik Classes are an invaluable resource for students seeking a comprehensive understanding of this important topic. Our solutions have been meticulously crafted to provide clarity and support in mastering the concepts and principles of states of matter.

By engaging with our solutions, students can develop a strong foundation in the three states of matter: solids, liquids, and gases. The step-by-step explanations, examples, and practice questions enable students to enhance their problem-solving skills and gain confidence in tackling a variety of questions related to states of matter.

Swastik Classes aims to make the learning experience enjoyable and accessible. Our NCERT Solutions simplify complex concepts, foster critical thinking, and stimulate curiosity in the captivating field of states of matter. We strive to instill a lifelong love for the subject and empower students to excel academically.

With the assistance of our experienced educators and subject matter experts, students can effectively grasp the theoretical aspects of states of matter while also developing their analytical and reasoning abilities. The solutions are designed to cater to the needs of students and educators alike, offering a reliable and comprehensive guide for academic success.

Swastik Classes is committed to providing top-quality educational resources, and our NCERT Solutions for Class 11 Chemistry Chapter 5 – States of Matter exemplify our dedication to excellence. Let Swastik Classes be your trusted companion on your educational journey, helping you unlock the mysteries of different states of matter and achieve academic excellence.

1. What is the kinetic molecular theory? The kinetic molecular theory is a model that explains the behavior of matter in terms of particles in motion. According to this theory, particles in a gas, liquid, or solid are constantly in motion, and their behavior is governed by factors such as temperature, pressure, and intermolecular forces.
2. What are intermolecular forces and how do they affect the properties of matter? Intermolecular forces are attractive forces that exist between molecules. These forces determine the physical properties of matter such as boiling point, melting point, and viscosity. Strong intermolecular forces result in higher boiling and melting points, while weak intermolecular forces lead to lower boiling and melting points.
3. What are the different types of intermolecular forces? The main types of intermolecular forces are:
• London dispersion forces (also known as Van der Waals forces): These are weak forces that exist between all molecules, regardless of polarity.
• Dipole-dipole interactions: These forces occur between polar molecules and are stronger than London dispersion forces.
• Hydrogen bonding: It is a special type of dipole-dipole interaction that occurs when hydrogen is bonded to a highly electronegative atom (such as oxygen, nitrogen, or fluorine).
4. What are the gas laws and how do they describe the behavior of gases? Gas laws are mathematical relationships that describe the behavior of gases under different conditions. Some important gas laws include Boyle’s law (pressure-volume relationship), Charles’s law (volume-temperature relationship), and Avogadro’s law (volume-amount relationship). These laws help us understand how gases respond to changes in temperature, pressure, and volume.
5. How can the behavior of gases be described using the ideal gas equation? The ideal gas equation, also known as the equation of state for an ideal gas, is PV = nRT. It relates the pressure (P), volume (V), amount of substance (n), gas constant (R), and temperature (T) for an ideal gas. This equation provides a useful framework for understanding and calculating the properties and behavior of gases.