## NCERT Solutions Class 11 Chemistry Chapter 5 – Free PDF Download

**According to the latest update on the CBSE Syllabus 2022-23, this chapter has been removed.*

**NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter **are available on this page to help students get a clear idea of the basic concepts. Detailed and step-by-step solutions according to the latest CBSE Syllabus 2022-23 are provided to every question listed in the NCERT textbook can be found here. This chapter mainly provides complete information regarding the different laws, kinds of thermal energy and intermolecular forces. NCERT Solutions for Class 11 Chemistry are designed with the main aim of helping the Class 11 students face the board exams without fear.

Solving the exercise questions using the solutions will provide a strong grip on the basic concepts. By regular practise, students will be able to analyse their areas of weakness and work on them for a better score in the term-II Class 11 Chemistry exam. The solutions contain authentic information by the experts after conducting a wide research on the important concepts. Also, these NCERT Solutions can be downloaded in a PDF format (for free) by clicking the download button provided above.

## Download NCERT Solutions Class 11 Chemistry Chapter 5 PDF

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

“**States of Matter**” is the fifth chapter in the NCERT Class 11 Chemistry textbook. This chapter contains several fundamental concepts related to intermolecular forces and how they affect the physical state of a substance. It also deals with several other important concepts associated with the liquid and gaseous states of matter. This is the reason why “states of matter” is regarded by many as one of the most important chapters in the NCERT Class 11 Chemistry textbook.

The topics covered in this chapter include intermolecular forces and thermal energy, gas laws, the ideal gas equation, the kinetic theory of gases, the deviation of gases from ideal behaviour, liquefaction of gases and the liquid state. The NCERT Solutions for Class 11 Chemistry (Chapter 5) provided on this page deal with the following types of questions:

- Numerical problems based on Boyle’s law, Charles’s law, Gay-Lusscac’s law and Avogadro’s law.
- Numerical problems on calculating partial pressure.
- Questions on critical temperature and pressure.
- Questions on Van der Waals forces and other types of intermolecular forces.

### Subtopics of Class 11 Chemistry Chapter 5: States of Matter

- Intermolecular Forces
- Dispersion Forces Or London Forces
- Dipole-dipole Forces
- Dipole-induced Dipole Forces
- Hydrogen Bond

- Thermal Energy
- Intermolecular Forces Vs Thermal Interactions
- The Gaseous State Ex
- The Gas Laws Ex
- Boyle’s Law (Pressure-volume Relationship)
- Charles’ Law (Temperature-volume Relationship)
- Gay Lussac’s Law (Pressure-temperature Relationship)
- Avogadro Law (Volume – Amount Relationship)

- Ideal Gas Equation
- Density And Molar Mass Of A Gaseous Substance
- Dalton’s Law Of Partial Pressures

- Kinetic Molecular Theory Of Gases
- Behaviour Of Real Gases: Deviation From Ideal Gas Behaviour
- Liquefaction Of Gases
- Liquid State
- Vapour Pressure
- Surface Tension and Viscosity.

## Access the answers of NCERT Solutions for Class 11 Chemistry Chapter 5

*Q1. 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:**

Initial pressure, P_{1} = 1 bar

Initial volume, V_{1} = 500

Final volume, V_{2} = 200

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

Acc. Boyle’s law,

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

P_{2 }=

=

= 2.5 bar

∴ the *minimum pressure required to compress is * 2.5 bar.

**Q2. 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:**

Initial pressure, P_{1} = 1.2 bar

Initial volume, V_{1} = 120 mL^{ }

Final volume, V_{2} = 180 mL

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

According to the Boyle’s law,

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

P_{2 }=

=

= 0.8 bar

Therefore, the min pressure required is 0.8 bar.

*Q3. Using the equation of state pV=nRT; show that at a given temperature density of a gas is proportional to gas pressure p.*

**Answer:**

The equation of state is given by,

pV = nRT ……..(1)

Where, p = pressure

V = volume

N = number of moles

R = Gas constant

T = temp

Replace n with

Where, m = mass

M = molar mass

But,

Where, d = density

Therefore, from equation (2), we get

d = (

d

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

*Q4. At 0°C, the density of a certain oxide of a gas at 2 bar is the same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?*

**Answer:**

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

d =

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

**Where, M _{1 }= mass of the oxide**

**p _{1 }= pressure of the oxide**

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

Where, M_{2 }= mass of the oxide

p_{2 }= pressure of the oxide

Acc to the question,

d_{1 }= d_{2}

Therefore,

Given:

Molecular mass of nitrogen,

Now,

=

=

= 70 g/mol

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

*Q5. The 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 the same temperature the pressure becomes 3 bar. Find a relationship between their molecular masses.*

**Answer:**

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

Where

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

Where,

From equation (1),

From equation (2),

Where,

Now, from equation (3) and (4),

Given,

Substituting these values in equation (5),

4

Therefore, the relationship between the molecular masses of X and Y is,

4

*Q6. 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? *

**Answer:**

The reaction of aluminum with caustic soda is as given below:

2Al + 2NaOH + 2H_{2}O _{2} + 3H_{2}

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

Therefore, 0.15 g Al gives:

= _{2}

= 186.67 mL of H_{2}

At Standard Temperature Pressure,

Let the volume of dihydrogen be

Now,

=

= 202.98 mL

= 203 mL

Hence, 203 mL of dihydrogen will be released.

* *

*Q7. 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 dm ^{3} flask at 27 °C? *

**Answer:**

It is known that,

p =

For methane (CH_{4}),

= ^{3 }= ^{3}

= 5.543 ×

For carbon dioxide (CO_{2}),

=

= 2.771 ×

Total pressure exerted by the mixture can be calculated as:

p =

= (5.543 ×

= 8.314 ×

* *

*Q8. 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?*

**Answer:**

Let the partial pressure of

Now,

It is known that,

= 0.4 bar

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

Now,

= 1.4 bar

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

= 0.4 + 1.4

= 1.8 bar

* *

*Q9. The 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? *

**Answer:**

Given,

d_{1} = 5.46 g/dm^{3}

p_{1 }= 2 bar

T_{1} =

p_{2 }= 1 bar

T_{2 }= 273 K

d_{2 }= ?

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

d =

d_{2} =

=

= 3 g dm^{-3}

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

*Q10. 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?*

**Answer:**

Given,

p = 0.1 bar

V = 34.05 mL = 34.05 × 10^{-3}

R = 0.083 bar ^{ }K^{-1} mol^{-1}

T =

The no of moles (n) can be calculated using the ideal gas equation as:

pV = nRT

n =

=

= 5.01 × 10^{-4} mol

Therefore, molar mass of phosphorus =

= 125 g mol ^{-1}

*Q11. A student forgot to add the reaction mixture to the container at $27^{\circ}$ C but instead, he placed the container on the flame. After a lapse of time, he came to know about his mistake, and using a pyrometer he found the temp of the container $477^{\circ}$ C. What fraction of air would have been expelled out?*

**Answer:**

Let the volume of the container be V.

The volume of the air inside the container at

Now,

V_{1} = V

T_{1} = _{2} = ?

T_{2} =

Acc to Charles’s law,

=

= 2.5 V

Therefore, volume of air expelled out

= 2.5 V – V = 1.5 V

Hence, fraction of air expelled out

=

=

*Q12. Calculate the temperature of 4.0 mol of a gas occupying 5 dm ^{3} at 3.32 bar. (R = 0.083 bar dm^{3} K^{–1} mol^{–1}). *

Given,

N= 4.0 mol

V = 5

p = 3.32 bar

R = 0.083 bar ^{ }K^{-1} mol^{-1}

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

pV = nRT

T =

=

= 50 K

Therefore, the required temp is 50 K.

*Q13. Calculate the total number of electrons present in 1.4 g of dinitrogen gas.*

**Answer:**

Molar mass of dinitrogen (N_{2}) = 28 g mol^{-1}

Thus, 1.4 g of N_{2 }

=

= 0.05 mol

= 0.05 × 6.02 × 10^{23} no. of molecules

= 3.01 × 10^{23 }no. of molecules

Now, 1 molecule of N_{2 }has 14 electrons.

Therefore, 3.01 × 10^{23 }molecules of N_{2} contains,

= 14 × 3.01 × 1023

= 4.214 × 10^{23} electrons

*Q14. How much time would it take to distribute one Avogadro number of wheat grains, if 10 ^{10} grains are distributed each second?*

**Answer:**

Avogadro no. = 6.02 × 10^{23}

Therefore, time taken

=

= 6.02 × 10^{13} s

=

= 1.909 × 10^{6} years

Therefore, the time taken would be 1.909 × 10^{6} years.

*Q15. Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 dm ^{3} at 27°C. R = 0.083 bar dm^{3} K^{–1} mol^{–1}*

**Answer:**

Given:

Mass of O_{2 }= 8 g

No. of moles

=

= 0.25 mole

Mass of H_{2} = 4 g

No. of moles

=

= 2 mole

Hence, total no. of moles in the mixture

= 0.25 + 2

= 2.25 mole

Given:

V = 1

n = 2.25 mol

R = 0.083 bar ^{-1} mol^{-1}

T =

Total pressure :

pV = nRT

p =

=

= 56.025 bar

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

*Q16. 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 dm^{3} K^{–1} mol^{–1})*

**Answer:**

Given:

r = 10 m

Therefore, volume of the balloon

= ^{3}

=

= 4190.5 m^{3} (approx.)

Therefore, the volume of the displaced air

= 4190.5 × 1.2 kg

= 5028.6 kg

Mass of helium,

=

Where, M = 4 × 10^{-3} kg mol^{-1}

p = 1.66 bar

V = volume of the balloon

= 4190.5 m^{3}

R = 0.083 0.083 bar ^{-1} mol^{-1}

T = 27 °C = 300 K

Then,

m =

= 1117.5 kg (approx.)

Now, total mass with helium,

= (100 + 1117.5) kg

= 1217.5 kg

Therefore, pay load,

= (5028.6 – 1217.5)

= 3811.1 kg

Therefore, the pay load of the balloon is 3811.1 kg.

*Q17. Calculate the volume occupied by 8.8 g of CO _{2} at 31.1°C and 1 bar pressure. R = 0.083 bar dm^{3} K^{–1} mol^{–1}.*

**Answer:**

pVM = mRT

V =

Given:

m = 8.8 g

R = 0.083 bar ^{-1} mol^{-1.}

T = 31.1 °C = 304.1 K

M = 44 g

p = 1 bar

Thus, Volume (V),

=

= 5.04806 L

= 5.05 L

Therefore, the volume occupied is 5.05 L.

*Q18. 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?.*

**Answer:**

Volume,

V =

=

Let M be the molar mass of the unknown gas.

Volume occupied by the unknown gas is,

=

=

According to the ques,

M =

= 40 g mol^{-1}

Therefore, the molar mass of the gas is 40 g mol^{-1}

*Q19. A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen.*

**Answer:**

Let the weight of dihydrogen be 20 g.

Let the weight of dioxygen be 80 g.

No. of moles of dihydrogen (nH_{2}),

=

= 10 moles

No. of moles of dioxygen (nO_{2}),

=

= 2.5 moles

Given:

p_{total} = 1 bar

Therefore, partial pressure of dihydrogen (pH_{2}),

= _{total}

=

= 0.8 bar

Therefore, the partial pressure of dihydrogen is 0.8 bar.

*Q20. What will be the SI unit for the quantity $\frac{ pV^{ 2 }T^{ 2 } }{ n }$?*

**Answer:**

SI unit of pressure, p =

SI unit of volume, V =

SI unit of temp, T = K

SI unit of number of moles, n = mol

Hence, SI unit of

=

=

*Q21. In terms of Charles’ law explain why $-273^{\circ}$ C is the lowest possible temp.*

**Answer:**

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

Charles found that for all gases, at any given pressure, the graph of volume vs temperature (in Celsius) is a straight line and on extending to zero volume, each line intercepts the temperature axis at – 273.15°C.

We can see that the volume of the gas at – 273.15°C will be zero. This means that gas will not exist. In fact, all the gases get liquified before this temperature is reached.

* *

*Q22. 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?*

**Answer:**

If the critical temperature of a gas is higher then it is easier to liquefy. That is the intermolecular forces of attraction among the molecules of gas are directly proportional to its critical temp.

Therefore, in CO_{2} intermolecular forces of attraction are stronger.

* *

*Q23. Explain the physical significance of Van der Waals parameters?*

**Answer:**

After accounting for pressure and volume corrections, the van der Waals equation is

The van der Waals constants or parameters are a and b.

The relevance of a and b is crucial here:

The magnitude of intermolecular attractive forces within the gas is measured by the value of ‘a,’ which is independent of temperature and pressure.

The volume occupied by the molecule is represented by ‘b,’ while the total volume occupied by the molecules is represented by ‘nb.’

Also Access |

NCERT Exemplar for class 11 chemistry Chapter 5 |

CBSE Notes for class 11 chemistry Chapter 5 |

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We at BYJU’S strive to create high-quality content and to make it available to all knowledge seekers. The **NCERT Solutions for Class 11 Chemistry** (Chapter 5) of the CBSE syllabus provided here have been carefully constructed while keeping the needs of CBSE Class 11 students in mind. These NCERT Solutions are crafted to be concept focused and can, therefore, be used by students to revise the key topics in the chapter. The Chemistry subject experts who created these solutions have provided simple yet detailed solutions to every NCERT intext and exercise question. Students can also reach out to our support team to clear any concept related doubts they may face.

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### Are NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter suitable for CBSE students?

**NCERT Solutions**have been prescribed for years as a complete source of information to CBSE students, to develop their analytical skills. They have proven to be essential for learning the syllabus and developing the confidence that is required to face their board exams. The NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter explain the steps with precision, without missing out on essential aspects of solving a question.