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

NCERT solutions for class 11 chemistry chapter 5 States of Matter- The matter is existing in 3 physical states which are solid, liquid and gas. In NCERT solutions for class 11 chemistry chapter 5 States of matter, you will deal with the two states of matter namely liquid state and gaseous state, the laws governing the behaviour of ideal gases and properties associated with liquids. In solutions of NCERT class 11 chemistry chapter 5 States of Matter, there are 23 questions in the exercise. The CBSE NCERT solutions for class 11 chemistry chapter 5 States of Matter are prepared and solved by subject experts. These NCERT solutions will help you in preparation of class 11 final examination as well as in the preparation of various competitive exams like NEET, JEE, BITSAT etc.

NCERT solutions for class 11 chemistry chapter 5 States of Matter is an important chapter for class 11 students because the topics of this chapter are the basics to the topics to be studied in class 12. The important topics of this chapter are intermolecular forces, thermal energy, the gaseous state, ideal gas equation, kinetic molecular theory of gases, liquid State and more. After completing the solutions of NCERT  class 11 chemistry chapter 5 states of matter students will be able to explain the existence of different states of matter, explain the laws governing the behaviour of ideal gases; able to apply gas laws in real life situations; describe the conditions required for liquefaction of gases and also able to differentiate between vapours and gaseous state.

Some common characteristics of three forms of matter solid liquid and gas are summarised below-

 S.No. Gases Liquids Solids 1 No definite shape No definite shape Definite shape 2 Indefinite volume Definite volume Definite volume 3 Compressible Slightly compressible Nearly compressible 4 Low density Intermediate density High density

## NCERT solutions for class 11 chemistry chapter 5 States of Matter- Exercise Questions

Given the volume of air to be compressed is  at pressure to  at .

So, as the temperature remains constant at .

We have Boyle's law where,

To calculate the final pressure .

We have,

Therefore, the minimum pressure required is .

Given volume of air to be transferred from the capacity of  vessel at pressure to  vessel at

So, as the temperature remains constant at .

We have Boyle's law where,

To calculate the final pressure .

We have,

Therefore, the pressure required is .

Given ideal gas equation ;

So, at given fixed temperature T. we have the

And we know

so, we get

But at constant Temperature, we have .

Given the condition :

Temperature

for oxide gas and   for dinitrogen.

As Density of gas is represented by,

And given that density is the same for both the gases at the same temperature condition.

So, we have   (as R is constant}

or,

or,

Given Pressure of 1g of an ideal gas A at  is .

When 2g of another ideal gas B is introduced in the same flask at the same temperature,

The pressure becomes .

.

We can assume the molecular masses of A and B be  respectively.

The ideal gas equation,

So, we have    and

Therefore,

or,

Hence the relation between the two gases is .

We have the chemical reaction:

(Where dihydrogen is being produced.)

So, at  and  pressure, the volume of gas that will be released when  of aluminium reacts.

As we can see from the reaction equation that 2 moles of aluminium produce 3 moles of dihydrogen.

i.e.,   reacts to give

At STP condition

and  ,

of Al gives  of .

Therefore  of Al will give  of

i.e.,  of .

At STP condition:

and  ,

Now,  ,

Then we assume the volume of dihydrogen be  at the pressure

(Since 1 bar = 0.987atm)  and temperature

Therefore, 203mL of dihydrogen will be released.

Given a mixture of 3.2g of methane and 4.4g of carbon dioxide contained a  flask at .

So, the pressure exerted by the mixture will be

The pressure exerted by the Methane gas,

The pressure exerted by the Carbon dioxide gas,

So, the total pressure exerted

And in terms of SI units, we get,

Pressure of the gas mixture will be the sum of partial pressure of  gas and partial pressure of  gas.

So, calculating the partial pressures of each gas.

Partial pressure of  in a 1L volume vessel.

As the temperature remains constant,

or     or

Now, calculating the partial pressure of  gas in 1L vessel.

or

Therefore the total pressure

Given the density of a gas is equal to  at  and at 2bar pressure.

Density , for the same gas at different temperatures and pressures we can apply,

Here,  ,

then at STP, we have

or

Given,

The volume of phosphorus vapour   which weights about  .

Temperature

Pressure

Hence we apply the ideal gas equation,

, where the number of moles will be

Therefore the molar mass is

Assume the round-bottomed flask has volume

so, the volume of air in the flask at  .

To find out how much air has been expelled out, we use Charle's Law:

Therefore

Or, Fraction of air expelled

Question

We have the ideal gas equation,

Pressure, p=3.32bar

Volume, V=5

Number of moles, n=4 mol

Gas constant R= 0.083

Temperature,  T=?

so, temperature, T=50K

Dinitrogen   has a molar mass .

Given 1.4g of  gas.

That means

number of molecules.

And as 1 molecule of  has 14 electrons.

So,  molecules of  has  electrons.

Given that  grains are distributed each second.

Time taken to distribute  grains = 1second.

Time taken to distribute one Avogadro number of wheat grains.

Given the mass of oxygen gas and mass of hydrogen gas.

Molar mass of

Therefore

Molar mass of

Therefore

Total number of moles of mixture

and

So, Ideal gas equation;

or

Therefore Total pressure is

(Density of and ).

The payload can be defined as:

(Mass of the displaced air - Mass of the balloon)

Given the radius of the balloon, r = 10 m.

Mass of the balloon, m = 100 kg.

Therefore, the volume of the balloon will be:

Now, the volume of the air displaced:

The mass of the air displaced :

Let  be the mass of helium gas filled into the balloon, then

Or,

approximately.

The balloon is filled with He with total mass of

The payload of the balloon will be:

Given the mass of carbon dioxide is 8.8grams at  and at 1 bar pressure.

So, the number of moles of

Now, the pressure of

Given

also, the

The Ideal gas equation;

So volume occupied by 8.8 g of carbon dioxide is 5.048L

Let the molar mass of the gas be ,

Then, given that  of a gas at  occupied the same volume as

of  at

So, we have the relation,    and

Therefore

wihch gives,

Or,

Or,

A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen.

So, let us assume the weight of dihydrogen in a total mixture weight of  be  and dioxygen be .

Then, the number of moles of  will be,  and the number of moles of .

And given that the total pressure of the mixture is .

Then we have a partial pressure of ,

Therefore, the partial pressure of  is .

Question

Given quantity, ,

The SI units of each factors are,

For pressure p is .

For Volume V is .

For Temperaute T is.

For number of moles, n is mol.

Therefore we have for the quantity

The SI unit is

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.

The critical temperature we know is the highest temperature at which liquid exists above it is gas.

Given that the critical temperature of Carbon dioxide and Methane are.

Higher is the critical temperature of a gas, easier is its liquefaction.

So, as the critical temperature increases the gas is now easier to liquefaction.

That means the intermolecular forces of attraction between the molecules of a gas are directly proportional to its Critical temperature.

Hence, Carbon dioxide has stronger intermolecular forces than Methane.

Question

The equation of van der Waals after taking into account the corrections for pressure and volume,

Where a and b are called van der Waals constants or parameters.

Here the significance of a and b is important:

Value of 'a' is a measure of the magnitude of intermolecular attractive forces within the gas and is independent of temperature and pressure.

And the value of 'b' is the volume occupied by the molecule and 'nb' is the total volume occupied by the molecules.

## NCERT solutions for class 11 chemistry

 Chapter 1 NCERT solutions for class 11 chemistry chapter 1 Some Basic Concepts of Chemistry Chapter-2 CBSE NCERT solutions for class 11 chemistry chapter 2 Structure of Atom Chapter-3 Solutions of NCERT class 11 chemistry chapter 3 Classification of Elements and Periodicity in Properties Chapter-4 NCERT solutions for class 11 chemistry chapter 4 Chemical Bonding and Molecular Structure Chapter-5 CBSE NCERT solutions for class 11 chemistry chapter 5 States of Matter Chapter-6 Solutions of NCERT class 11 chemistry chapter 6 Thermodynamics Chapter-7 NCERT solutions for class 11 chemistry chapter 7 Equilibrium Chapter-8 CBSE NCERT solutions for class 11 chemistry chapter 8 Redox Reaction Chapter-9 Solutions of NCERT class 11 chemistry chapter 9 Hydrogen Chapter-10 NCERT solutions for class 11 chemistry chapter 10 The S-Block Elements Chapter-11 CBSE NCERT solutions for class 11 chemistry chapter 11 The P-Block Elements Chapter-12 Solutions of NCERT class 11 chemistry chapter 12 Organic chemistry- some basic principles and techniques Chapter-13 NCERT solutions for class 11 chemistry chapter 13 Hydrocarbons Chapter-14 CBSE NCERT solutions for class 11 chemistry chapter 14 Environmental Chemistry

## NCERT solutions for class 11 subject wise

 NCERT solutions for class 11 biology Solutions of NCERT class 11 maths CBSE NCERT solutions for class 11 chemistry NCERT solutions for class 11 physics

## Benefits of NCERT solutions for class 11 chemistry chapter 5 States of Matter

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