NCERT Solutions Class 11 Chemistry Chapter 5 – Free PDF Download
*According to the latest update on the CBSE Syllabus 2023-24, 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 2023-24 are provided for every question listed in the NCERT textbook. 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 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. Through regular practice, students will be able to analyse their areas of weakness and work on them for a better score in the Class 11 Chemistry exam. The solutions contain authentic information by the experts after conducting wide research on the important concepts. Also, these NCERT Solutions can be downloaded in a PDF format (for free) by clicking the download button.
NCERT Solutions for Class 11 Chemistry Chapter 5: States of Matter
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, the 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 in 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 dm3 of air at 1 bar to 200 dm3 at 30°C?
Answer:
Initial pressure, P1Â = 1 bar
Initial volume, V1Â = 500
Final volume, V2Â = 200
As the temperature remains the same, the final pressure (P2) can be calculated with the help of Boyle’s law.
According to Boyle’s law,
P1V1Â = P2V2
P2Â =
=
= 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, P1Â = 1.2 bar
Initial volume, V1Â = 120 mLÂ
Final volume, V2Â = 180 mL
As the temperature remains the same, the final pressure (P2) can be calculated with the help of Boyle’s law.
According to Boyle’s law,
P1V1Â = P2V2
P2Â =
=
= 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 = Temperature
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 temperature, 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:
The density (d) of the substance at temperature (T) can be given by
d =
Now, the density of oxide (d1) is given as
Where M1 = Mass of the oxide
p1 = Pressure of the oxide
The density of dinitrogen gas (d2) is given as,
Where M2 = Mass of the oxide
p2 = Pressure of the oxide
According to the question,
d1 = d2
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 equations (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 aluminium 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 aluminium reacts?
Answer:
The reaction of aluminium with caustic soda is given below.
2Al + 2NaOH + 2H2O
At Standard Temperature Pressure (273.15 K and 1 atm), 54 g ( 2 × 27 g) of Al gives 3 ×22400 mL of H2.
Therefore, 0.15 g Al gives
=
= 186.67 mL of H2
At Standard Temperature Pressure,
Let the volume of dihydrogen be
Now,
=
= 202.98 mL
= 203 mL
Hence, 203 mL of dihydrogen will be released.
Â
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 dm3 flask at 27 °C?
Answer:
It is known that,
p =
For methane (CH4),
=
= 5.543 ×
For carbon dioxide (CO2),
=
= 2.771 ×
The total pressure exerted by the mixture can be calculated as
p =
= (5.543 ×
= 8.314 ×
Â
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 O2 in the container be
Now,
= 1.4 bar
The total pressure of the gas mixture in the container can be obtained as
= 0.4 + 1.4
= 1.8 bar
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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,
d1 = 5.46 g/dm3
p1 = 2 bar
T1 =
p2 = 1 bar
T2 = 273 K
d2 = ?
The density ( d2 ) of the gas at STP can be calculated using the equation,
d =
d2 =
=
= 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 × 10-4 mol
Therefore, molar mass of phosphorus =
= 125 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,
V1 = V
T1 =
T2 =
According to Charles’s law,
=
= 2.5 V
Therefore, the volume of air expelled out
= 2.5 V – V = 1.5 V
Hence, the fraction of air expelled out
=
=
Q12. 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).
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 (N2) = 28 g mol-1
Thus, 1.4 g of N2
=
= 0.05 mol
= 0.05 × 6.02 × 1023 no. of molecules
= 3.01 × 1023 no. of molecules
Now, 1 molecule of N2 has 14 electrons.
Therefore, 3.01 × 1023 molecules of N2 contains
= 14 × 3.01 × 1023
= 4.214 × 1023 electrons
Q14. How much time would it take to distribute one Avogadro number of wheat grains, if 1010 grains are distributed each second?
Answer:
Avogadro no. = 6.02 × 1023
Therefore, the time taken
=
= 6.02 × 1013 s
=
= 1.909 × 106 years
Therefore, the time taken would be 1.909 × 106 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 dm3 at 27°C. R = 0.083 bar dm3 K–1 mol–1
Answer:
Given:
Mass of O2 = 8 g
No. of moles
=
= 0.25 mole
Mass of H2 = 4 g
No. of moles
=
= 2 mole
Hence, the 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 dm3 K–1 mol–1)
Answer:
Given:
r = 10 m
Therefore, the volume of the balloon
=
=
= 4190.5 m3 (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 m3
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, payload
= (5028.6 – 1217.5)
= 3811.1 kg
Therefore, the payload 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 dm3 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 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.
The volume occupied by the unknown gas is
=
=
According to the question,
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 (nH2)
=
= 10 moles
No. of moles of dioxygen (nO2)
=
= 2.5 moles
Given:
ptotal = 1 bar
Therefore, partial pressure of dihydrogen (pH2)
=
=
= 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 the number of moles, n = mol
Hence, SI unit of
=
=
Q21. In terms of Charles’ law explain why
Answer:
Charles’ law states that pressure remains constant, and 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.
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Q22. The 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 temperature.
Therefore, in CO2 intermolecular forces of attraction are stronger.
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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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