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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 121-150 of 185 questions

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121 Medium
121 of 185
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.
122 Medium
122 of 185
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.
123 Medium
123 of 185
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.
124 Medium
124 of 185
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.
125 Medium
125 of 185
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.
126 Medium
126 of 185
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.
127 Medium
127 of 185
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.
128 Medium
128 of 185
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.
129 Medium
129 of 185
Understanding +4 | -1
The attraction that non-polar molecules have for each other is primarily caused by:
View answer and explanation
Correct option: D
Van der Waals’ forces
Option D (Van der Waals’ forces) is correct because non-polar molecules interact through weak, temporary dipoles induced by electron fluctuations.
130 Medium
130 of 185
Application +4 | -1
The molar mass of a gas is 50 g / mol.
The density of the gas at critical
temperature and critical pressure of 30
atm is 125 g / L. What is the critical
temperature of the gas?
[Take : R = 0.08 L atm mol−1{}^{ - 1} K−1{}^{ - 1}]
View answer and explanation
Correct option: A
400
Option A (400) is correct because the critical temperature can be calculated using the critical compressibility factor and the ideal gas law.
131 Medium
131 of 185
Understanding +4 | -1
In a given mixture of gases which do not react with one another, the ratio of partial pressure of each component to the total pressure is equal to its
View answer and explanation
Correct option: C
Mole fraction
Option C (Mole fraction) is correct because, according to Dalton's law of partial pressures, the ratio of the partial pressure of a gas to the total pressure is equal to its mole fraction.
132 Medium
132 of 185
Application +4 | -1
Under identical conditions of temperature the density of a gas AA is three times that of gas BB while molecular mass of gas BB is twice that of AA. The ratio of pressures of AA and BB will be:
View answer and explanation
Correct option: A
6
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.
133 Medium
133 of 185
Understanding +4 | -1
Which is valid at absolute zero?
View answer and explanation
Correct option: B
KEKE of the molecules becomes zero and the molecular motion also becomes zero
The correct answer is B (KEKE of the molecules becomes zero and the molecular motion also becomes zero) because at absolute zero (0 Kelvin), all molecular motion ceases, and the kinetic energy becomes zero.
134 Medium
134 of 185
Understanding +4 | -1
Average speed of molecules of a gas sample is equal to ________ times their RMS speed
View answer and explanation
Correct option: C
0.9213
Option C (0.9213) is correct because the average speed of gas molecules is approximately 0.9213 times their root-mean-square (RMS) speed, based on the kinetic theory of gases.
135 Medium
135 of 185
Understanding +4 | -1
Statement I: Standard boiling point of the liquid is less than the normal boiling point.

Statement II: 1 bar pressure is slightly less than 1 atm pressure.
View answer and explanation
Correct option: B
Both Statement I and Statement II are correct and Statement II is not the correct explanation of Statement I.
Option B is correct because both statements are correct, but the second statement (pressure difference) does not explain the first statement (boiling point difference); the boiling point difference is a direct consequence of the pressure difference.
136 Medium
136 of 185
Application +4 | -1
The volume of oxygen gas required at STP conditions for the complete combustion of 15 cc of methane gas at 27∘C27^\circ C and 750 mm pressure
View answer and explanation
Correct option: D
26.94 cc
Option D (26.94 cc) is correct because it involves converting the given conditions to STP, then using stoichiometry to find the volume of oxygen required for complete combustion of methane.
137 Medium
137 of 185
Application +4 | -1
The average kinetic energy of an ideal gas per molecule in SI units at 250C{25^0}C will be
View answer and explanation
Correct option: B
6.17×10−21J6.17 \times {10^{ - 21}}\mathrm{J}
Option B (6.17×10−21J6.17 \times {10^{ - 21}}\mathrm{J}) is correct because it is the average kinetic energy calculated using the formula 32kT\frac{3}{2}kT with the correct Boltzmann constant and temperature.
138 Medium
138 of 185
Application +4 | -1
Partial pressure of H2{H_2}, if flask containing 2 g of H2{H_2} and 32 g of SO2S{O_2} is:
View answer and explanation
Correct option: C
2Ptotal3\frac{{2{P_{\mathrm{total}}}}}{3}
Option C (2Ptotal3\frac{{2{P_{\mathrm{total}}}}}{3}) is correct because the partial pressure of a gas is the product of the total pressure and its mole fraction.
139 Medium
139 of 185
Application +4 | -1
Let the most probable velocity of hydrogen molecules at a temperature of t∘C\mathrm{t}^{\circ} \mathrm{C} be V0\mathrm{V}_0. When the temperature is raised to (2t+273)∘C(2 \mathrm{t}+273)^{\circ} \mathrm{C} the new rms velocity is (suppose all the molecules dissociate into atoms at latter temperature)
View answer and explanation
Correct option: B
6 V0\sqrt{6} \mathrm{~V}_0
The correct answer is B (6 V0\sqrt{6} \mathrm{~V}_0) because the new RMS velocity can be found by relating it to the initial most probable velocity, considering the dissociation of hydrogen molecules into atoms.
140 Medium
140 of 185
Application +4 | -1
Density of an ideal gas at 2 atm2 \mathrm{~atm} and 600 K600 \mathrm{~K} is 2 g/L2 \mathrm{~g} / \mathrm{L}. Calculate relative density of this gas with respect to Ne(g)\mathrm{Ne}(\mathrm{g}) at same conditions : (Given : R=1/12 atm L/mol.KR=1 / 12 \mathrm{~atm} \mathrm{~L} / \mathrm{mol} . \mathrm{K} )
View answer and explanation
Correct option: A
2.52.5
Option A (2.52.5) is correct because the relative density is calculated using the ideal gas law and the definition of relative density, resulting in approximately 2.5.
141 Medium
141 of 185
Remembering +4 | -1
Boyle’s temperature and inversion temperature are related as
View answer and explanation
Correct option: C
Ti=2TbT_i=2T_b
Option C (Ti=2TbT_i=2T_b) is correct because the inversion temperature (TiT_i) is twice the Boyle's temperature (TbT_b).
142 Medium
142 of 185
Application +4 | -1
The ratio of rate of diffusion of helium and methane under identical conditions of pressure and temperature is:
View answer and explanation
Correct option: B
2
Option B (2) is correct because the ratio of diffusion rates is inversely proportional to the square root of the molar masses, resulting in a ratio of 2.
143 Medium
143 of 185
Application +4 | -1
The temperature of 20 L of nitrogen was increased from 10 K to 30 K at a constant pressure. Change in volume will be
View answer and explanation
Correct option: B
40L40\mathrm{ }L
Option B (40L40\mathrm{ }L) is correct because the change in volume is calculated using Charles's Law, resulting in a 40 L increase.
144 Medium
144 of 185
Analyzing +4 | -1
0.5 mole of each of H2,SO2{\mathrm{H}_2},\mathrm{S}{\mathrm{O}_2} and CH4\mathrm{C}{\mathrm{H}_4} are kept in a container. A hole was made in the container. After 3 h, the order of partial pressures in the container will be
View answer and explanation
Correct option: B
pSO2>pCH4>pH2p\mathrm{S}{\mathrm{O}_2} > p\mathrm{C}{\mathrm{H}_4} > p{\mathrm{H}_2}
Option B (pSO2>pCH4>pH2p\mathrm{S}{\mathrm{O}_2} > p\mathrm{C}{\mathrm{H}_4} > p{\mathrm{H}_2}) is correct because the rate of diffusion is inversely proportional to the square root of the molecular mass. Therefore, the gas with the highest molecular mass will have the lowest diffusion rate and hence the highest partial pressure after 3 hours.
145 Medium
145 of 185
Application +4 | -1
An open flask containing air is heated from 300 k to 500 k. What percentage of air will be escaped to the atmosphere, if pressure is keeping constant
View answer and explanation
Correct option: B
40
Option B (40) is correct because using Charles's Law, the volume increases proportionally with temperature, and the percentage of air escaped is calculated based on the change in volume.
146 Medium
146 of 185
Understanding +4 | -1
What is the reason for unusual high b.p. of water?
View answer and explanation
Correct option: A
Due to the presence of H+{\mathrm{H}^ + } and OH−{\mathrm{OH}^ - }ions in water
The correct answer is A (Due to the presence of H+{\mathrm{H}^ + } and OH−{\mathrm{OH}^ - }ions in water) because the high boiling point of water is primarily due to hydrogen bonding between water molecules.
147 Medium
147 of 185
Application +4 | -1
A gaseous mixture was prepared by taking
equal number of moles of Helium and Neon. If
the total pressure of the mixture was found to
be 10 atm, the partial pressure of Helium in the
mixture is ----
View answer and explanation
Correct option: B
5 atm
Option B (5 atm) is correct because the partial pressure of a gas in a mixture is equal to its mole fraction multiplied by the total pressure. Since the mole fraction of Helium is 0.5, its partial pressure is 5 atm.
148 Medium
148 of 185
Application +4 | -1
An open flask containing air is heated from 300 k to 500 k. What percentage of air will be escaped to the atmosphere, if pressure is keeping constant
View answer and explanation
Correct option: B
40
Option B (40) is correct because using Charles's Law, the volume increases proportionally with temperature, and the percentage of air escaped is calculated based on the change in volume.
149 Medium
149 of 185
Understanding +4 | -1
If saturated vapours are compressed slowly at constant temperature to half the initial volume, the vapour pressure will
View answer and explanation
Correct option: D
Remains unchanged
Option D (Remains unchanged) is correct because at constant temperature, the vapor pressure of a saturated vapor remains constant regardless of volume changes.
150 Medium
150 of 185
Understanding +4 | -1
Statement I: The order of viscosities of hexane, water glycerol is hexane << water << glycerol.

Statement II: Hexane has weakest intermolecular force and glycerol the strongest.
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 (Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I) is correct because viscosity is directly related to intermolecular forces, with stronger forces leading to higher viscosity.
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