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 
Question 5.1 What will be the minimum pressure required to compress of air at 1 bar to at ?
Answer:
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 .
Answer:
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 .
Question 5.3 Using the equation of state show that at a given temperature density of a gas is proportional to gas pressure p.
Answer:
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 .
Answer:
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,
Answer:
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 .
Answer:
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.
Answer:
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,
Answer:
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
Question 5.9 Density of a gas is found to be at at 2 bar pressure. What will be its density at STP ?
Answer:
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
Question 5.10 of phosphorus vapour weighs 0.0625 g at and 0.1 bar pressure. What is the molar mass of phosphorus?
Answer:
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
Answer:
Assume the roundbottomed 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 5.12 Calculate the temperature of of a gas occupying at 3.32 bar.
Answer:
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
Question 5.13 Calculate the total number of electrons present in 1.4 g of dinitrogen gas.
Answer:
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.
Question 5.14 How much time would it take to distribute one Avogadro number of wheat grains, if 1010 grains are distributed each second ?
Answer:
Given that grains are distributed each second.
Time taken to distribute grains = 1second.
Time taken to distribute one Avogadro number of wheat grains.
Question 5.15 Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of at 27°C. .
Answer:
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
Answer:
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:
Question 5.17 Calculate the volume occupied by 8.8 g of at 31.1°C and 1 bar pressure. .
Answer:
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
Answer:
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,
Answer:
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 5.20 What would be the unit for the quantity ?
Answer:
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
Question 5.21 In terms of Charles’ law explain why –273 °C is the lowest possible temperature.
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.
Answer:
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 5.23 Explain the physical significance of van der Waals parameters.
Answer:
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.
Chapter 1 
NCERT solutions for class 11 chemistry chapter 1 Some Basic Concepts of Chemistry 
Chapter2 
CBSE NCERT solutions for class 11 chemistry chapter 2 Structure of Atom 
Chapter3 

Chapter4 
NCERT solutions for class 11 chemistry chapter 4 Chemical Bonding and Molecular Structure 
Chapter5 
CBSE NCERT solutions for class 11 chemistry chapter 5 States of Matter 
Chapter6 
Solutions of NCERT class 11 chemistry chapter 6 Thermodynamics 
Chapter7 
NCERT solutions for class 11 chemistry chapter 7 Equilibrium 
Chapter8 
CBSE NCERT solutions for class 11 chemistry chapter 8 Redox Reaction 
Chapter9 

Chapter10 
NCERT solutions for class 11 chemistry chapter 10 The SBlock Elements 
Chapter11 
CBSE NCERT solutions for class 11 chemistry chapter 11 The PBlock Elements 
Chapter12 

Chapter13 
NCERT solutions for class 11 chemistry chapter 13 Hydrocarbons 
Chapter14 
CBSE NCERT solutions for class 11 chemistry chapter 14 Environmental Chemistry 
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5.21 In terms of Charles’ law explain why –273 °C is the lowest possible temperature.
5.7 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 flask at ?
5.3 Using the equation of state show that at a given temperature density of a gas is proportional to gas pressure p.