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JEE • CHEMISTRY • STATES OF MATTER

States of matter practice · Medium

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 1-30 of 185 questions

1 Medium

1 of 185

Application +4 | -1

A quantity of hydrogen gas occupies a volume of

30 \mathrm{~mL}

at a certain temperature and pressure. What volume would half this mass of hydrogen occupy at triple the absolute temperature and the pressure were one-ninth that of the original gas?

View answer and explanation

Correct option: A

405 \mathrm{~mL}

Option A (405 mL) is correct because it is calculated using the combined gas law, considering the changes in mass, temperature, and pressure.

2 Medium

2 of 185

Understanding +4 | -1

Consider the following statements for kinetic molecular theory of gases
a. Particles of a gas move in all possible directions in straight lines.
b. At any particular time, different particles in the gas have different kinetic energies.
c. Collisions of gas molecules are perfectly inelastic in nature.
The correct statement(s) is/are

View answer and explanation

Correct option: C

a and b only

Option C (a and b only) is correct because gas particles move in straight lines and different particles have different kinetic energies at a given time, but collisions are perfectly elastic.

3 Medium

3 of 185

Understanding +4 | -1

The relative strength of interionic/intermolecular forces in decreasing order is

View answer and explanation

Correct option: B

ion-ion > ion-dipole > dipole-dipole

Option B is correct because ion-ion interactions are the strongest, followed by ion-dipole, and then dipole-dipole interactions due to the magnitude of charges involved.

4 Medium

4 of 185

Remembering +4 | -1

Compressibility factor of an ideal gas is

View answer and explanation

Correct option: B

Equal\mathrm{ }to\mathrm{ }1\;

Option B (

Equal\mathrm{ }to\mathrm{ }1\;

) is correct because the compressibility factor (Z) for an ideal gas is defined as 1.

5 Medium

5 of 185

Analyzing +4 | -1

A: Hot air balloon rises up by displacing the denser air of the atmosphere.
R: The given mass of a gas becomes less denser.

View answer and explanation

Correct option: A

If both A and R are correct and R is the correct explanation of A

Option A is correct because the hot air balloon rises due to the heated air inside being less dense than the surrounding cooler air, as described by the relationship between density and temperature.

6 Medium

6 of 185

Understanding +4 | -1

Pressure cooker reduces cooking time because

View answer and explanation

Correct option: C

Boiling point of water is elevated

Option C (Boiling point of water is elevated) is correct because a pressure cooker increases the pressure, which elevates the boiling point of water, leading to faster cooking.

7 Medium

7 of 185

Application +4 | -1

Under identical conditions of temperature the density of a gas

A

is three times that of gas

B

while molecular mass of gas

B

is twice that of

A

. The ratio of pressures of

A

and

B

will be:

View answer and explanation

Correct option: A

Option A (6) is correct because using the ideal gas law in terms of density, the ratio of pressures can be found to be 6.

8 Medium

8 of 185

Understanding +4 | -1

Statement I: The Joule-Thomson coefficient for an ideal gas is zero

Statement II: There are no intermolecular attractive forces between molecules in an ideal gas.

View answer and explanation

Correct option: A

Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I

Option A is correct because both statements are true, and the absence of intermolecular forces in an ideal gas directly leads to a zero Joule-Thomson coefficient.

9 Medium

9 of 185

Application +4 | -1

If Z is the compressibility factor, vanderwaal’s equation at low pressure can be written as

View answer and explanation

Correct option: B

Z = 1 - \frac{9}{{V\,RT}}

Option B (

Z = 1 - \frac{9}{{V\,RT}}

) is correct because at low pressure, the van der Waals equation simplifies to this form by considering the attractive forces between molecules.

10 Medium

10 of 185

Remembering +4 | -1

The van der Waals’ equation for real gas is:

View answer and explanation

Correct option: C

P = \frac{{nRT}}{{V - nb}} - \frac{{a{n^2}}}{{{V^2}}}

Option D (All of the above) is correct because all the given equations are valid representations of the van der Waals equation for real gases.

11 Medium

11 of 185

Analyzing +4 | -1

A: At very high pressures, compressibility factor is greater than 1.
R: At very high pressure, '

b

' can be neglected in van der Waal's gas equation.

View answer and explanation

Correct option: C

If A is correct but R is incorrect

Option D (The complexes [NiCl4]2- and [Ni(CN)4]2- differ in primary valencies of nickel) is correct because the primary valency (oxidation state) of Ni is +2 in both complexes, so they do NOT differ in primary valencies.

12 Medium

12 of 185

Application +4 | -1

The rate of diffusion of two gases

x

and

y

is 10 ml/s
and 40 ml/s respectively then ratio of their molar
masses is :-

View answer and explanation

Correct option: D

0.667361111

Option D (16:01) is correct because the ratio of the rates of diffusion is inversely proportional to the square root of the ratio of their molar masses, as described by Graham's Law.

13 Medium

13 of 185

Understanding +4 | -1

A comparison is made at Standard Temperature and Pressure (STP) of 0.50 mole of

\mathrm{H}_2(g)

and 1 mole of

\mathrm{He}(g)

. The two gases will

View answer and explanation

Correct option: A

have equal average molecular kinetic energies

Option A (have equal average molecular kinetic energies) is correct because the average kinetic energy of a gas depends only on temperature, and both gases are at the same temperature (STP).

14 Medium

14 of 185

Application +4 | -1

Two flasks

A

and

B

of equal volumes maintained at temperatures

300 \mathrm{~K}

and

600 \mathrm{~K}

contain equal mass of

\mathrm{H}_2

and

\mathrm{CH}_4

respectively. The ratio of total translational kinetic energy of gas in flask

A

to that in flask

B

View answer and explanation

Correct option: C

Option C (4) is correct because the ratio of kinetic energies is directly proportional to the ratio of (moles * temperature), and after calculation, it equals 4.

15 Medium

15 of 185

Analyzing +4 | -1

Assertion (A): The value of van der Waals' constant '

a

' is higher for ammonia than for nitrogen.

Reason (R): Intermolecular hydrogen bonding is present in ammonia.

View answer and explanation

Correct option: A

Assertion is correct, reason is correct; reason is a correct explanation for assertion

Option A is correct. The van der Waals constant 'a' reflects the strength of intermolecular forces, and ammonia exhibits hydrogen bonding, which is a stronger intermolecular force than the London dispersion forces present in nitrogen.

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16 Medium

16 of 185

Understanding +4 | -1

During the evaporation of liquid

View answer and explanation

Correct option: B

The temperature of the liquid will fall

Option B is correct. During evaporation, the molecules with the highest kinetic energy escape from the liquid's surface, thus lowering the average kinetic energy and therefore the temperature of the remaining liquid.

17 Medium

17 of 185

Remembering +4 | -1

The interaction energy of London forces between two particles is proportional to

{r^x}

, where

r

is the distance between the particles. The value of

x

View answer and explanation

Correct option: C

-6

Option C (-6) is correct because the interaction energy due to London dispersion forces is inversely proportional to the sixth power of the distance between the particles, hence

{r^{-6}}

18 Medium

18 of 185

Application +4 | -1

300 \mathrm{~K}

, the density of a certain gaseous molecule at 2 bar is double to that of dinitrogen

\left(\mathrm{N}_2\right)

at 4 bar. The molar mass of gaseous molecule is :

View answer and explanation

Correct option: C

112 \mathrm{~g} \mathrm{~mol}^{-1}

Option C (

112 \mathrm{~g} \mathrm{~mol}^{-1}

) is correct because it is calculated using the ideal gas law and the relationship between density, pressure, molar mass, and temperature.

19 Medium

19 of 185

Application +4 | -1

If the pressure is halved and absolute temperature doubled the volume of the gas will be:

View answer and explanation

Correct option: A

Option A (4) is correct because using the combined gas law, if the pressure is halved and the absolute temperature is doubled, the volume will be 4 times the original volume.

20 Medium

20 of 185

Application +4 | -1

At 400 K, the root mean square (rms) speed of a gas

X

(molecular weight = 40) is equal to the most probable speed of gas

Y\,\,at\,\,60\,\,K

. The molecular weight of the gas Y is

View answer and explanation

Correct option: B

Option B (4) is correct because it is derived from equating the root mean square speed of gas X to the most probable speed of gas Y and solving for the molecular weight of gas Y.

21 Medium

21 of 185

Application +4 | -1

Which of the following expression represent correctly the variation of density of an ideal gas with change in temperature?

View answer and explanation

Correct option: A

{d_2} = \frac{{{P_2}{T_1}{d_1}}}{{{P_1}{T_2}}}

Option A (

{d_2} = \frac{{{P_2}{T_1}{d_1}}}{{{P_1}{T_2}}}

) is correct because it correctly relates density, pressure, and temperature for an ideal gas based on the ideal gas law.

22 Medium

22 of 185

Remembering +4 | -1

Van der Waals’ forces are applied to:

View answer and explanation

Correct option: C

Mixture of gases

Option C (Mixture of gases) is correct because Van der Waals' forces are intermolecular forces that exist between all types of molecules, including those in a mixture of gases.

23 Medium

23 of 185

Understanding +4 | -1

A helium atoms is two times heavier than a hydrogen molecule. At 298 K, the average kinetic energy of a helium atom is

View answer and explanation

Correct option: D

Same as that of a hydrogen molecule

Option D is correct because the average kinetic energy of a gas depends only on the temperature and is independent of the mass of the gas molecules.

24 Medium

24 of 185

Application +4 | -1

Equal volumes of

{\mathrm{H}_2}

and

\mathrm{C}{\mathrm{l}_2}

are mixed. How will the volume of the mixture change after the reaction?

View answer and explanation

Correct option: A

Unchanged

Option A (Unchanged) is correct because the reaction between equal volumes of H2 and Cl2 produces HCl, and the total volume of gases remains the same.

25 Medium

25 of 185

Application +4 | -1

The density of a gas

A

is twice that of a gas

B

at the same temperature. The molecular weight of gas

B

is thrice that of

A

. The ratio of the pressures acting on

A

and

B

will be

View answer and explanation

Correct option: A

\frac{1}{6}

Option A (

\frac{1}{6}

) is correct because the ratio of pressures is inversely proportional to the ratio of densities and directly proportional to the ratio of molecular weights.

26 Medium

26 of 185

Application +4 | -1

The relative rates of diffusion of

{\mathrm{U}^{235}}{\mathrm{F}_6}

and

{\mathrm{U}^{238}}{\mathrm{F}_6}

are

View answer and explanation

Correct option: A

1.0043

Option A (1.0043) is correct because it is calculated using Graham's law of diffusion, considering the molar masses of the isotopes.

27 Medium

27 of 185

Application +4 | -1

At 400 K, the root mean square (rms) speed of a gas

X

(molecular weight = 40) is equal to the most probable speed of gas

Y\,\,at\,\,60\,\,K

. The molecular weight of the gas Y is

View answer and explanation

Correct option: B

Option B (4) is correct because equating the RMS speed of gas X to the most probable speed of gas Y and solving for the molecular weight of Y yields 4.

28 Medium

28 of 185

Analyzing +4 | -1

Assertion (A): Wet air is heavier than dry air.

Reason (R): The density of dry air is more than density of water.

View answer and explanation

Correct option: D

Assertion is incorrect, reason is correct

Wet air is lighter than dry air because water vapor (H2O) has a lower molar mass than the average molar mass of dry air (mostly N2 and O2). The reason is correct that dry air is denser than water, but this doesn't explain why wet air is lighter.

29 Medium

29 of 185

Application +4 | -1

{p_A}

and

{p_B}

are the vapour pressure of pure liquid components

A

and

B

respectively of an ideal binary solution. If

{\chi _A}

represents the mole fraction of component

A,

the total pressure of the solution will be

View answer and explanation

Correct option: D

{p_B} + {\chi _A}\left( {{p_A} - {p_B}} \right)

Option D (

{p_B} + {\chi _A}\left( {{p_A} - {p_B}} \right)

) is correct because it represents the total pressure of an ideal binary solution according to Raoult's law.

30 Medium

30 of 185

Application +4 | -1

Two identical cylinders contain helium at

2.5atm

and argon at

1atm

respectively. if both the gases are filled in one of the cylinders, the pressure would be:

View answer and explanation

Correct option: A

3.5\,\,\mathrm{atm}

Option A (3.5 atm) is correct because when gases are mixed in the same volume, the total pressure is the sum of the individual pressures (Dalton's Law).

Revision notes and key takeaways

Review STATES OF MATTER in your textbook or class notes, then use this set to check your understanding.

Key Takeaways

✓ This set contains 185 single-correct-answer MCQs on STATES OF MATTER.
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