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JEE • CHEMISTRY • STATES OF MATTER

States of matter practice · Mixed difficulty

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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 151-180 of 262 questions

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151 Medium
151 of 262
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.
152 Hard
152 of 262
Application +4 | -1
An open vessel containing air is heated from 300 K to 400 K. The fraction of air originally present which goes out of it is:
View answer and explanation
Correct option: B
14\frac{1}{4}
Option B (14\frac{1}{4}) is correct because it represents the fraction of air that escapes the vessel when heated from 300 K to 400 K, calculated using the ideal gas law.
153 Medium
153 of 262
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.
154 Medium
154 of 262
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.
155 Medium
155 of 262
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.
156 Medium
156 of 262
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.
157 Medium
157 of 262
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.
158 Medium
158 of 262
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.
159 Medium
159 of 262
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.
160 Medium
160 of 262
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.
161 Medium
161 of 262
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.
162 Medium
162 of 262
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.
163 Easy
163 of 262
Application +4 | -1
Relative density of a volatile substance with respect to O2O_2 is 8. Its molecular weight would be
View answer and explanation
Correct option: C
256
Option C (256) is correct because the molecular weight is calculated using the relative density with respect to O2O_2.
164 Medium
164 of 262
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).
165 Easy
165 of 262
Application +4 | -1
The minimum pressure required to compress 200 dm3200 \mathrm{~dm}^3 of air at 2 atm2 \mathrm{~atm} to 50dm350 \mathrm{dm}^3 at 25∘C25^{\circ} \mathrm{C} is
View answer and explanation
Correct option: C
8 atm8 \mathrm{~atm}
Option C (8 atm) is correct because it is the pressure calculated using Boyle's Law, which states that for a fixed amount of gas at constant temperature, the pressure and volume are inversely proportional.
166 Medium
166 of 262
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.
167 Hard
167 of 262
Application +4 | -1
To an evacuated vessel with movable piston under external pressure of 1 atm, 0.1 mole of He and 1.0 mole of an unknown compound (vapour pressure 0.68 atm at 0  ∘C0\,\,^\circ C ) are introduced. Considering the ideal gas behaviour, the total volume (in litre) of the gases at 0 ∘C0\,^\circ C  is close to
View answer and explanation
Correct option: C
7
Option C (7) is correct because the partial pressure of He is calculated, and then using the ideal gas law, the volume is found to be approximately 7L.
168 Medium
168 of 262
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.
169 Medium
169 of 262
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.
170 Easy
170 of 262
Remembering +4 | -1
The compressibility factor for an ideal gas is
View answer and explanation
Correct option: B
1
Option B (1) is correct because the compressibility factor (Z) for an ideal gas is always equal to 1, indicating ideal behavior.
171 Medium
171 of 262
Remembering +4 | -1
“One gram molecule of a gas at N.T.P occupies 22.4 lit” This fact was derived from
View answer and explanation
Correct option: B
Avogadro’s hypothesis
Option B (Avogadro’s hypothesis) is correct because Avogadro's hypothesis states that equal volumes of all gases, at the same temperature and pressure, have the same number of molecules, leading to the 22.4 L molar volume at NTP.
172 Medium
172 of 262
Understanding +4 | -1
With increase of pressure, the mean free path
View answer and explanation
Correct option: A
Decreases
Option A (Decreases) is correct because as pressure increases, the molecules are closer together, leading to more frequent collisions and a shorter mean free path.
173 Medium
173 of 262
Application +4 | -1
If temperature of 1 mole of gas is increased by 50 ∘C50\,^\circ C calculate the change in kinetic energy of the system
View answer and explanation
Correct option: A
623.25J623.25\mathrm{ }J
Option A (623.25J623.25\mathrm{ }J) is correct because it is the change in kinetic energy calculated using the formula KE = (3/2)RT.
174 Medium
174 of 262
Remembering +4 | -1
The temperature at which a real gas obeys the ideal gas laws over a wide range of pressure is
View answer and explanation
Correct option: B
Boyle temperature
Option B (Boyle temperature) is correct because Boyle temperature is the temperature at which a real gas obeys the ideal gas law over a wide range of pressures.
175 Easy
175 of 262
Understanding +4 | -1
The interaction force present between H2\mathrm{H}_2 molecules are
View answer and explanation
Correct option: C
London forces
Option C (London forces) is correct because H2 is a nonpolar molecule, and the only intermolecular forces present are London dispersion forces.
176 Medium
176 of 262
Application +4 | -1
Find the ratio of moles of SO2\mathrm{SO}_2 to CH4\mathrm{CH}_4 after fifth diffusion steps if their initially mole ratio is 8:1.8: 1.
View answer and explanation
Correct option: A
14\frac{1}{4}
Option A (14\frac{1}{4}) is correct because it is calculated using Graham's law of diffusion after 5 diffusion steps.
177 Easy
177 of 262
Understanding +4 | -1
Vander waal constant “b” is very high for the following gas
View answer and explanation
Correct option: A
Cl2C{l_2}
Option A (Cl2C{l_2}) is correct because the van der Waals constant 'b' is directly proportional to the size/volume of the gas molecule.
178 Medium
178 of 262
Application +4 | -1
The relative rates of diffusion of U235F6{\mathrm{U}^{235}}{\mathrm{F}_6} and U238F6{\mathrm{U}^{238}}{\mathrm{F}_6} are
View answer and explanation
Correct option: A
1.0043
Option A (1.0043) is correct because the relative rates of diffusion are inversely proportional to the square root of their molar masses.
179 Medium
179 of 262
Understanding +4 | -1
The van der Waals’ equation for a real gas is given by the formula (p+n2aV2)(V−nb)=nRT,wherep,V,Tandn\left( {p + \frac{{{n^2}a}}{{{V^2}}}} \right)\left( {V - nb} \right) = nRT,\mathrm{where}p,V,T\mathrm{and}n are the pressure, volume, temperature and the number of moles of the gas. Which one is the correct interpretation for the parameter a?
View answer and explanation
Correct option: C
The parameter aa accounts for intermolecular interaction’s present in the molecule
Option C (The parameter aa accounts for intermolecular interaction’s present in the molecule) is correct because the 'a' parameter in the van der Waals equation accounts for the attractive intermolecular forces between gas molecules.
180 Medium
180 of 262
Application +4 | -1
A mixture of Ne and Ar kept in a closed vessel at 250 K250 \mathrm{~K} has a total K.E. =3 kJ=3 \mathrm{~kJ}. The total mass of Ne\mathrm{Ne} and Ar\mathrm{Ar} is 30 g30 \mathrm{~g}. Find mass % of Ne\mathrm{Ne} in gaseous mixture at 250 K250 \mathrm{~K}.
View answer and explanation
Correct option: D
28.3
Option D (28.3) is correct because it is the mass percentage of Ne calculated from the total kinetic energy and total mass of the mixture.
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