NCERT Solutions for Class 11 Chemistry: Chapter 5 (States of Matter) are provided on this page for the perusal of Class 11 chemistry students. Detailed, stepbystep solutions to every question listed in Chapter 5 of the NCERT Class 11 chemistry textbook can be found here. Also, these NCERT solutions can be downloaded in a PDF format (for free) by clicking the download button provided above.
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NCERT Solutions for Chemistry – Class 11, 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, GayLusscac’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
 Dipoledipole Forces
 Dipoleinduced Dipole Forces
 Hydrogen Bond
 Thermal Energy
 Intermolecular Forces Vs Thermal Interactions
 The Gaseous State Ex

The Gas Laws Ex
 Boyle’s Law (Pressurevolume Relationship)
 Charles’ Law (Temperaturevolume Relationship)
 Gay Lussac’s Law (Pressuretemperature 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.
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
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:
=
= 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}),
=
= 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 H2 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
T =
The no of moles (n) can be calculated using the ideal gas equation as:
pV = nRT
n =
=
= 5.01 × 10^{5} mol
Therefore, molar mass of phosphorus =
= 1247.5 g mol^{1}
Q11. A student forgot to add the reaction mixture to the container at
Answer:
Let the volume of the container be V.
The volume of the air inside the container at
Now,
V_{1} = V
T_{1} =
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
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
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
=
=
= 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
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 CO2 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
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 (n_{H2}),
=
= 10 moles
No. of moles of dioxygen (n_{O2}),
=
= 2.5 moles
Given:
p_{total} = 1 bar
Therefore, partial pressure of dihydrogen (p_{H2}),
=
=
= 0.8 bar
Therefore, the partial pressure of dihydrogen is 0.8 bar.
Q20. What will be the SI unit for the quantity
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
Answer:
According to Charles’ law
At constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temp.
It was found that for all gasses (at any given pressure), the plot of volume vs. temp. (in
If we extend the line to zero volume, then it intersects the tempaxis at
Therefore, it can be said that
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 temp 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:
The physical significance of ‘a’:
The magnitude of intermolecular attractive forces within gas is represented by ‘a’.
The physical significance of ‘b’:
The volume of a gas molecule is represented by ‘b’.
Also Access 
NCERT Exemplar for class 11 chemistry Chapter 5 
CBSE Notes for class 11 chemistry Chapter 5 
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