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JEE • CHEMISTRY • STATES OF MATTER

States of matter practice · Medium

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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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91 Medium
91 of 185
Application +4 | -1
A perfect gas of a given mass is heated first in a small vessel and then in a large vessel, such that their volume remains unchanged. The P−TP - T curves are:
View answer and explanation
Correct option: D
Linear with different slopes
The correct answer is Linear with different slopes because the P−TP-T relationship is linear, and the slope depends on the volume.
92 Medium
92 of 185
Understanding +4 | -1
Following properties will decrease with increase in temperature except
View answer and explanation
Correct option: D
Vapour pressure
Option D (Vapour pressure) is correct because vapor pressure increases with increasing temperature due to increased kinetic energy of molecules.
93 Medium
93 of 185
Understanding +4 | -1
If the assumption that there is no force of attraction between the molecules of a gas is correct, what will be the consequences?
View answer and explanation
Correct option: B
The gases will never liquefy when cooled and compressed
Option B is correct because if there is no attraction between gas molecules, they will not come close enough to liquefy, even when cooled and compressed.
94 Medium
94 of 185
Application +4 | -1
X mLX \mathrm{~mL} of H2\mathrm{H}_2 gas effuse through a hole in a container in 5 s5 \mathrm{~s}. The time taken for the effusion of the same volume of the gas specified below under ideal conditions is
View answer and explanation
Correct option: B
20 s,O220 \mathrm{~s}, \mathrm{O}_2
Option B (20 s,O220 \mathrm{~s}, \mathrm{O}_2) is correct because the effusion rate is inversely proportional to the square root of the molar mass.
95 Medium
95 of 185
Application +4 | -1
A gas of unknown identity effuses at the rate of 83.3  mLs−183.3\,\,\mathrm{mL}{\mathrm{s}^{ - 1}} in an effusion apparatus in which carbon dioxide effuses at the rate of 102  mLs−1102\,\,\mathrm{mL}{\mathrm{s}^{ - 1}}. Calculate molar mass of the unknown gas
View answer and explanation
Correct option: B
65.97  g mol−165.97\,\,\mathrm{g}\,\mathrm{mo}{\mathrm{l}^{ - 1}}
Option B (65.97  g mol−165.97\,\,\mathrm{g}\,\mathrm{mo}{\mathrm{l}^{ - 1}}) is correct because it is calculated using Graham's law of effusion, relating effusion rates to molar masses.
96 Medium
96 of 185
Application +4 | -1
Equal masses of CH4C{H_4} and O2{O_2} are mixed in a container at 25∘ C25^\circ \,C. Fraction of total pressure exerted by O2{O_2} is
View answer and explanation
Correct option: B
13\frac{1}{3}
Option B (13\frac{1}{3}) is correct because the partial pressure of a gas is proportional to its mole fraction in the mixture.
97 Medium
97 of 185
Application +4 | -1
In two separate bulbs containing ideal gases AA and BB respectively, the density of gas AA is twice that of gas BB while molecular weight of gas AA is half that of gas BB at the same temperature, pressure ratio PA/PB{P_A}/{P_B} will be:
View answer and explanation
Correct option: C
4
Option C (4) is correct because the pressure ratio is directly proportional to the ratio of densities and inversely proportional to the ratio of molecular weights, resulting in a ratio of 4.
98 Medium
98 of 185
Application +4 | -1
4.0g4.0\mathrm{g} of argon has pressure PP and temperature T  KT\,\,\mathrm{K} in a vessel. On keeping the vessel at 50∘50^\circ higher temperature, 0.8g0.8\mathrm{g} of argon was given out to maintain the pressure PP. The original temperature was:
View answer and explanation
Correct option: C
200  K200\,\,\mathrm{K}
Option C (200  K200\;\mathrm{K}) is correct because it is derived from applying the ideal gas law under the given conditions of constant pressure and volume.
99 Medium
99 of 185
Analyzing +4 | -1
Assertion : Compressibility factor (Z)(Z) for non-ideal gases can be greater than 1 .

Reason : Non-ideal gases always exert higher pressure than expected.
View answer and explanation
Correct option: C
If Assertion is correct but Reason is incorrect.
The correct answer is C because the assertion is correct (Z can be > 1) but the reason is incorrect (non-ideal gases can exert lower pressure than expected).
100 Medium
100 of 185
Understanding +4 | -1
Correct order of van der Waals constant b of given gases is
View answer and explanation
Correct option: A
CO2>N2>O2>H2>He\mathrm{CO}_2>\mathrm{N}_2>\mathrm{O}_2>\mathrm{H}_2>\mathrm{He}
Option D (All of these) is correct because all the given expressions are valid representations of Kf, the cryoscopic constant.
101 Medium
101 of 185
Understanding +4 | -1
Dipole-Dipole interaction energy b/w polar molecule in solid depends on the radius of the molecule (r)\left( r \right) and it is directly proportional to
View answer and explanation
Correct option: D
1r3\frac{1}{{{r^3}}}
Option D (1r3\frac{1}{{{r^3}}}) is correct because the dipole-dipole interaction energy in solids is inversely proportional to the cube of the distance (radius) between the molecules.
102 Medium
102 of 185
Application +4 | -1
Under what conditions will a pure sample of an ideal gas not only exhibit a pressure of 1 atm but also a concentration of 1mol  litre−11\mathrm{mol}\,\,\mathrm{litr}{\mathrm{e}^{ - 1}} ? (  R=0.082(\;R = 0.082 litre atm mol−1deg⁡−1)\mathrm{mo}{\mathrm{l}^{ - 1}}{\deg ^{ - 1}})
View answer and explanation
Correct option: C
When T=12KT = 12K
Option C (When T=12KT = 12K) is correct because it satisfies the ideal gas law equation with the given conditions.
103 Medium
103 of 185
Application +4 | -1
A closed cylinder having volume 20 L20 \mathrm{~L} contains 56 g N256 \mathrm{~g} {~N}_2 and 16 g He16 \mathrm{~g} {~He} gases. If pressure exerted by the gaseous mixture is 8.2 atm8.2 \mathrm{~atm} then the temperature of the gas molecules will be
View answer and explanation
Correct option: B
333.3 K333.3 \mathrm{~K}
Option B (333.3 K) is correct because it is calculated using the ideal gas law with the given pressure, volume, and number of moles.
104 Medium
104 of 185
Understanding +4 | -1
If pressure of a fixed quantity of a gas is increased 4 times keeping the temperature constant, the r.m.s. velocity will be
View answer and explanation
Correct option: C
Same
Option C (Same) is correct because the root mean square (rms) velocity depends only on temperature and molar mass, and since the temperature is constant, the rms velocity remains the same.
105 Medium
105 of 185
Understanding +4 | -1
Which of the assumption of the kinetic molecular theory best explains the observation that a balloon collapses when exposed to liquid nitrogen (which is much colder than a cold winter day)?
View answer and explanation
Correct option: B
The velocity of gas molecules is proportional to their kelvin temperature
Option B is correct because the kinetic molecular theory states that the velocity of gas molecules is proportional to their Kelvin temperature, explaining why a balloon collapses when exposed to liquid nitrogen.
106 Medium
106 of 185
Application +4 | -1
2 moles of NH3(g){NH}_3(\mathrm{g}) and 1 mole of HCl(g){HCl}(\mathrm{g}) are taken in a container of capacity 8.21 lit at 300 K300 \mathrm{~K} to produce NH4Cl(s){NH}_4{Cl}(\mathrm{s}). Find the total pressure after the reaction.
View answer and explanation
Correct option: A
3 atm
Option D (2 atm) is incorrect. Option A (3 atm) is correct because it is the total pressure calculated after the reaction, considering the remaining NH3 gas.
107 Medium
107 of 185
Application +4 | -1
The ratio of the root mean square velocity of H2\mathrm{H}_2 at 50 K50 \mathrm{~K} and that of O2\mathrm{O}_2 at 800 K800 \mathrm{~K} is
View answer and explanation
Correct option: C
1
Option C (1) is correct because the ratio of RMS velocities is calculated using the formula VRMS∝TMV_{RMS} \propto \sqrt{\frac{T}{M}} and simplifies to 1.
108 Medium
108 of 185
Application +4 | -1
Active mass of 14 grams of Nitrogen at STP is 'X\mathrm{X}'. Active mass of 8 grams of oxygen at STP is 'Y\mathrm{Y}'. Ratio of X\mathrm{X} to Y\mathrm{Y} is
View answer and explanation
Correct option: A
1:11: 1
Option A (1:1) is correct because the active mass of any gas at STP is directly proportional to the number of moles, and the ratio of moles is proportional to the ratio of active masses.
109 Medium
109 of 185
Analyzing +4 | -1
S-I: Compressibility factor for hydrogen varies with pressure with positive slope at all pressures at 0∘C0^{\circ} \mathrm{C}.
S-II: Even at low pressures, repulsive forces dominate hydrogen gas at 0∘C0^{\circ} \mathrm{C}.
View answer and explanation
Correct option: A
Both statements are correct and statement-II is correct explanation of statement-I
Option A (Both statements are correct and statement-II is correct explanation of statement-I) is correct because the compressibility factor for hydrogen increases with pressure due to the dominance of repulsive forces.
110 Medium
110 of 185
Application +4 | -1
The density of a gas AA is 4 times to that of a gas BB If the molecular mass of AA is MM what is the molecular mass of BB
View answer and explanation
Correct option: A
M4\frac{M}{4}
Option A (M4\frac{M}{4}) is correct because the molecular mass of B is one-fourth of the molecular mass of A, derived from the ideal gas law and the given density relationship.
111 Medium
111 of 185
Analyzing +4 | -1
Assertion (A): Different gases at the same conditions of temperature and pressure have same root mean square speed.

Reason (R): Average kinetic energy of a gas is directly proportional to temperature in kelvin scale.
View answer and explanation
Correct option: D
Assertion is incorrect, reason is correct
The correct answer is D (Assertion is incorrect, reason is correct) because the root mean square speed depends on the molar mass of the gas, and the average kinetic energy is directly proportional to temperature.
112 Medium
112 of 185
Application +4 | -1
A bottle of dry NH3\mathrm{N}{\mathrm{H}_3} and bottle of dry HCl\mathrm{HCl} connected through a long tube are opened simultaneously at both ends, the white (NH4Cl)\left( {\mathrm{N}{\mathrm{H}_4}\mathrm{Cl}} \right) ring first formed will be:
View answer and explanation
Correct option: B
Near the HCl\mathrm{HCl} bottle
Option B (Near the HCl\mathrm{HCl} bottle) is correct because NH3\mathrm{N}{\mathrm{H}_3} diffuses faster than HCl due to its lower molecular weight, leading to the ring forming closer to the HCl bottle.
113 Medium
113 of 185
Application +4 | -1
A glass bulb of 2 L2 \mathrm{~L} capacity is filled by helium gas at 10 atm10 \mathrm{~atm} pressure. Due to a leakage the gas leaks out. What is the volume of gas leaked if the final pressure in container is 1 atm.1 \mathrm{~atm}.
View answer and explanation
Correct option: A
18 L{1 8} \mathrm{~L}
Option A (18 L{1 8} \mathrm{~L}) is correct because it is the volume of gas leaked calculated using Boyle's law.
114 Medium
114 of 185
Application +4 | -1
A quantity of gas is collected in a graduated tube over the mercury. The volume of gas at 18∘C18^{\circ} \mathrm{C} is 50.0 mL50.0 \mathrm{~mL} and the level of mercury in the tube is 100 mm100 \mathrm{~mm} above the outside mercury level. The barometer reads 750 torr. Hence, volume at STP is approximately
View answer and explanation
Correct option: B
40 mL40 \mathrm{~mL}
Option B (40 mL) is correct because the volume at STP is calculated using the combined gas law, accounting for the difference in mercury levels and converting to standard conditions.
115 Medium
115 of 185
Application +4 | -1
Equal masses of H2,O2\mathrm{H}_2, \mathrm{O}_2 and methane have been taken in a container of volume V\mathrm{V} at temperature 27∘C27^{\circ} \mathrm{C} in identical conditions. The ratio of the volumes of gases H2:O2\mathrm{H}_2: \mathrm{O}_2 : methane would be :
View answer and explanation
Correct option: C
16:1:216: 1: 2
Option C (16:1:216: 1: 2) is correct because the ratio of volumes is directly proportional to the number of moles, and the number of moles is inversely proportional to the molecular mass.
116 Medium
116 of 185
Application +4 | -1
Equal masses of CH4C{H_4} and H2{H_2} are mixed in an empty container at 25∘C25^\circ C. The fraction of the total pressure exerted by H2{H_2} is
View answer and explanation
Correct option: B
89\frac{8}{9}
Option B (89\frac{8}{9}) is correct because the mole fraction of H2 is calculated as moles of H2 divided by the total moles, and then used to find the partial pressure fraction.
117 Medium
117 of 185
Remembering +4 | -1
S.I units of gas constant R is
View answer and explanation
Correct option: C
8.314  J. K−1.mol−18.314\,\,J.\,{K^{ - 1}}.mo{l^{ - 1}}
Option C (8.314  J. K−1.mol−18.314\,\,J.\,{K^{ - 1}}.mo{l^{ - 1}}) is correct because it represents the value of the gas constant R in S.I. units (Joules per Kelvin per mole).
118 Medium
118 of 185
Analyzing +4 | -1
Assertion: The molecules of the dissolved gas present in a liquid gain kinetic energy as temperature is raised.

Reason : Gases tends to be more soluble in liquids as the temperature is raised.
View answer and explanation
Correct option: C
If Assertion is correct but Reason is incorrect.
Option C is correct because the assertion is true, but the reason is false. Gases become less soluble as temperature increases.
119 Medium
119 of 185
Application +4 | -1
At the top of the mountain the thermometer reads 0∘C0^{\circ} \mathrm{C} and the barometer reads 710 mm710 \mathrm{~mm} of Hg\mathrm{Hg}. At the bottom of the mountain the temperature is 30∘C30^{\circ} \mathrm{C} and pressure is 760 mm760 \mathrm{~mm} of Hg\mathrm{Hg}. Density of air at the top with that at the bottom is
View answer and explanation
Correct option: B
1.04:11.04: 1
The correct answer is B (1.04: 1) because the density of a gas is directly proportional to pressure and inversely proportional to temperature, allowing for a direct comparison between the top and bottom of the mountain.
120 Medium
120 of 185
Application +4 | -1
A flask contains 12 g12 \mathrm{~g} of a gas of relative molecular mass 120 at a pressure of 100 atm100 \mathrm{~atm} was evacuated by means of a pump until the pressure was 0.01 atm0.01 \mathrm{~atm}. Which of the following is the best estimate of the number of molecules left in the flask (N0=6×1023 mol−1)\left(\mathrm{N}_0=6 \times 10^{23} \mathrm{~mol}^{-1}\right).
View answer and explanation
Correct option: B
6×10186 \times 10^{18}
Option B (6×10186 \times 10^{18}) is correct because it is calculated using the ratio of pressures to find the number of moles remaining and then multiplying by Avogadro's number.
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