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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 91-120 of 262 questions

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91 Easy
91 of 262
Remembering +4 | -1
The temperature at which real gases obey the Boyle’s law over a wide range of pressure is called
View answer and explanation
Correct option: B
Boyle’s temperature
Option B (Boyle’s temperature) is correct because Boyle's temperature is the temperature at which a real gas obeys Boyle's law over a wide range of pressures.
92 Medium
92 of 262
Application +4 | -1
One mole of gas AA and three moles of a gas BB are placed in flask of volume 100 litres at 27∘C27^{\circ} \mathrm{C}. Calculate the total partial pressure of the gases in the mixture.
View answer and explanation
Correct option: C
0.985 atm0.985 \mathrm{~atm}
Option C (0.985 atm0.985 \mathrm{~atm}) is correct because it is calculated using the ideal gas law with the total number of moles of gas.
93 Medium
93 of 262
Understanding +4 | -1
A pressure cooker reduces cooking time for food because
View answer and explanation
Correct option: A
Boiling point of water involved in cooking is increased.
Option A is correct because a pressure cooker increases the boiling point of water, allowing food to cook at a higher temperature and thus reducing cooking time.
94 Medium
94 of 262
Application +4 | -1
For a Vander Waal's gas the volume of gas molecules are negligible. The molar volume of gas is 20 L/mol20 \mathrm{~L} / \mathrm{mol} at 300 K300 \mathrm{~K}. The compressibility factor of the gas is

[Given: a=96 L2⋅atm⋅mol−2{a}=96 \mathrm{~L}^2 \cdot \mathrm{atm} \cdot \mathrm{mol}^{-2} and R=0.08 L⋅atm⋅mol−1 K−1{R}=0.08 \mathrm{~L} \cdot \mathrm{atm} \cdot \mathrm{mol}^{-1} \mathrm{~K}^{-1}]
View answer and explanation
Correct option: A
0.8
Option A (0.8) is correct because the compressibility factor is calculated using the van der Waals equation, considering the given parameters.
95 Hard
95 of 262
Analyzing +4 | -1
Statement I: Reduced equation of state in terms of reduce: temperature θ\theta, reduced volume ϕ\phi and reduced pressure x can be applied to solid, gases and liquid.

Statement II: (π+3ϕ2)(3ϕ−1)=8θ\left(\pi+\frac{3}{\phi^2}\right)(3 \phi-1)=8 \theta. Equation is independent of p,Vp, V and TT as well as independent of vander Waals' constant aa and bb.
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 D (N2+\mathrm{N}_2^ + becomes diamagnetic) is correct because removing an electron from N2, which has an odd number of electrons in its molecular orbitals, will not result in a diamagnetic species.
96 Medium
96 of 262
Understanding +4 | -1
In a mixture of a light gas and a heavy gas in a closed container, the light gas will:
View answer and explanation
Correct option: B
Have a higher average speed per molecule than the heavy gas
Option B is correct because, according to the kinetic molecular theory, lighter gas molecules have a higher average speed at the same temperature.
97 Medium
97 of 262
Understanding +4 | -1
To which of the following gaseous mixture, the Dalton’s law of partial pressures will not apply
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Correct option: C
NO+O2NO + {O_2}
Option C is correct because NO and O2 react to form NO2, thus Dalton's law is not applicable to the mixture.
98 Easy
98 of 262
Understanding +4 | -1
Equal volume of different gases at any definite temperature and pressure have
View answer and explanation
Correct option: D
Equal molecules
Option D (Equal molecules) is correct because Avogadro's law states that equal volumes of all gases at the same temperature and pressure contain the same number of molecules.
99 Easy
99 of 262
Application +4 | -1
If the density of certain gas at 30∘C30^\circ C 768 torr is 1.35  Kg/m31.35\,\,Kg/{m^3} its density at STP would be
View answer and explanation
Correct option: A
1.48 Kg/m31.48\,Kg/{m^3}
Option A (1.48 Kg/m31.48\,Kg/{m^3}) is correct because it is calculated using the ideal gas law and the relationship between density, pressure, temperature, and molar mass.
100 Medium
100 of 262
Application +4 | -1
By what factor does the average velocity of a gaseous molecule increase when the temperature (in Kelvin) is doubled?
View answer and explanation
Correct option: A
1.4
Option A (1.4) is correct because the average velocity is proportional to the square root of the temperature, so doubling the temperature increases the average velocity by a factor of 2≈1.4\sqrt{2} \approx 1.4.
101 Medium
101 of 262
Application +4 | -1
Oxygen is present in 1 litre flask at a pressure of 7.6×10−10 mm7.6 \times 10^{-10} \mathrm{~mm} of Hg\mathrm{Hg}. Calculate the number of oxygen molecules in the flask at 0∘C0^{\circ} \mathrm{C}.
View answer and explanation
Correct option: B
2.7×10102.7 \times 10^{10}
Option B (2.7×10102.7 \times 10^{10}) is correct because it is the number of oxygen molecules calculated using the ideal gas law and Avogadro's number.
102 Medium
102 of 262
Remembering +4 | -1
The joule Thomson coefficient is zero at
View answer and explanation
Correct option: C
Inversion temperature
Option C (Inversion temperature) is correct because the Joule-Thomson coefficient is zero at the inversion temperature, indicating no temperature change upon expansion.
103 Medium
103 of 262
Analyzing +4 | -1
Order of intermolecular attractive forces.
View answer and explanation
Correct option: B
CH3−O−CH3<C2H5OH<H2OC{H_3} - O - C{H_3} < {C_2}{H_5}OH < {H_2}O
Option B (CH3−O−CH3<C2H5OH<H2OC{H_3} - O - C{H_3} < {C_2}{H_5}OH < {H_2}O) is correct because the strength of intermolecular forces increases with hydrogen bonding, which is strongest in water due to having two H atoms.
104 Medium
104 of 262
Application +4 | -1
15 g15 \mathrm{~g} of a gas present in 5 L5 \mathrm{~L} vessel exerts a pressure of 0.1 bar0.1 \, \mathrm{bar}. What is the root mean square speed of the gas molecules?
View answer and explanation
Correct option: C
100 m/s100 \mathrm{~m} / \mathrm{s}
Option C (100 m/s100 \text{{ m/s}}) is correct because the root mean square speed is calculated using the formula vrms=3Pdv_{rms} = \sqrt{\frac{{3P}}{{d}}}, where P is pressure and d is density.
105 Medium
105 of 262
Understanding +4 | -1
A balloon filled with N2O{\mathrm{N}_2}\mathrm{O} is pricked with a sharp point and quickly plunged into a tank of CO2C{O_2} under the same pressure and temperature. The balloon will:
View answer and explanation
Correct option: C
Remain unchanged in size
Option C (Remain unchanged in size) is correct because N2O and CO2 have nearly the same molar mass, resulting in similar diffusion rates under the same conditions.
106 Easy
106 of 262
Application +4 | -1
Calculate the weight of methane in a 9L cylinder at 16 atm and 27∘C27^\circ C
View answer and explanation
Correct option: C
93.6 g
The correct answer is 93.6 g because it is calculated using the ideal gas law, PV = nRT, to find the number of moles and then the mass.
107 Medium
107 of 262
Application +4 | -1
Ratio of average to most probable velocity is
View answer and explanation
Correct option: A
1.1281.128
Option A (1.1281.128) is correct because the ratio of average velocity to most probable velocity is approximately 1.128.
108 Medium
108 of 262
Application +4 | -1
Boyle&#39;s temperature of helium is, given van der Waals&#39; constant a=0.0351 atm L2 mol−2\mathrm{a}=0.0351 \mathrm{~atm} \mathrm{~L}^2 \mathrm{~mol}^{-2} and b=0.024 L mol−1b=0.024 \mathrm{~L} \mathrm{~mol}^{-1}
View answer and explanation
Correct option: A
18 K18 \mathrm{~K}
Option A (18 K18 \mathrm{~K}) is correct because it is calculated using the formula Tb=aRbT_b = \frac{a}{Rb} with the given van der Waals constants.
109 Medium
109 of 262
Application +4 | -1
The root mean square speed of the molecules of diatomic gas is uu. When the temperature is doubled, the molecules dissociates into two atoms. The new speed of the atom is:
View answer and explanation
Correct option: C
2u2u
Option C (2u2u) is correct because when the temperature is doubled and the diatomic molecule dissociates, the new RMS speed becomes twice the original RMS speed.
110 Medium
110 of 262
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 compressibility factor (Z) can be greater than 1 for non-ideal gases, but non-ideal gases do not always exert higher pressure than expected.
111 Medium
111 of 262
Application +4 | -1
Using van der waal's equation, calculate the constant, ' aa ' (atm Ltr⁡2 mole−2\operatorname{Ltr}^2 \mathrm{~mole}^{-2} ) when two moles of a gas confined in a four litre flask exerts a pressure of 11 atm11 \mathrm{~atm} at a temperature of 300 K300 \mathrm{~K}. The value of ' b\mathrm{b} ' is 0.05Litremole 0.05 \mathrm{Litre}^{\text {mole }}.
View answer and explanation
Correct option: A
6.5
Option A (6.5) is correct because it is the value of the van der Waals constant 'a' calculated using the given parameters in the van der Waals equation.
112 Hard
112 of 262
Analyzing +4 | -1
Consider 1 cm3c{m^3} sample of air at absolute temperature T0{T_0} at sea-level and another 1  cm31\,\,c{m^3} sample of air at a height where the pressure is one third atmosphere. The absolute temperature TT of the sample at the height is :
View answer and explanation
Correct option: D
Cannot be determined in terms of T0{T_0} from the above data
Option D is correct because without knowing the mass of the air samples, we cannot determine the relationship between the temperatures using the ideal gas law.
113 Hard
113 of 262
Application +4 | -1
A given mass of helium at STP is allowed to expand to 1.4 times its initial volume by changing the temperature and pressure, then its resultant density would be
View answer and explanation
Correct option: C
0.13 kg m−30.13 \mathrm{~kg} \mathrm{~m}^{-3}
Option C (0.13 kg m−30.13 \mathrm{~kg} \mathrm{~m}^{-3}) is correct because density is inversely proportional to volume. When the volume increases by 1.4 times, the density decreases by 1.4 times.
114 Easy
114 of 262
Understanding +4 | -1
The ratio of volume occupied by 1 mole O2\mathrm{O}_2 and 1 mole CO2\mathrm{CO}_2 under identical conditions of temperature and pressure is:
View answer and explanation
Correct option: A
1:11: 1
Option A (1:11: 1) is correct because, according to Avogadro's law, equal volumes of all gases at the same temperature and pressure contain the same number of molecules, thus the volume ratio is directly proportional to the mole ratio.
115 Medium
115 of 262
Application +4 | -1
The rates of diffusion of SO2,CO2,PCl3\mathrm{S}{\mathrm{O}_2},\mathrm{C}{\mathrm{O}_2},\mathrm{PC}{\mathrm{l}_3} and SO3\mathrm{S}{\mathrm{O}_3} are in the following order
View answer and explanation
Correct option: D
CO2>SO2>SO3>PCl3\mathrm{C}{\mathrm{O}_2} > S{\mathrm{O}_2} > S{\mathrm{O}_3} > PC{\mathrm{l}_3}
Option D (CO2>SO2>SO3>PCl3\mathrm{C}{\mathrm{O}_2} > S{\mathrm{O}_2} > S{\mathrm{O}_3} > PC{\mathrm{l}_3}) is correct because the rate of diffusion is inversely proportional to the square root of the molecular weight.
116 Easy
116 of 262
Analyzing +4 | -1
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: Amongst He,Ne,Ar\mathrm{He}, \mathrm{Ne}, \mathrm{Ar} and Kr;1g\mathrm{Kr} ; 1\mathrm{g} of activated charcoal adsorbs more of Kr\mathrm{Kr}.
Reason R: The critical volume Vc(cm3 mol–1)\mathrm{V_c(cm^3 \,mol^{–1}}) and critical pressure PcP_c (atm) is highest for Krypton but the compressibility factor at critical point ZcZ_c is lowest for Krypton.
In the light of the above statements, choose the correct answer from the options given below.
View answer and explanation
Correct option: B
A is true but R is false
The correct answer is B (A is true but R is false) because while Kr is adsorbed more, the given reason about critical volume, pressure, and compressibility factor is incorrect.
117 Hard
117 of 262
Understanding +4 | -1
The equation, [Pr+3Vr2][3Vr−1]=8Tr:\left[ {{P_r} + \frac{3}{{V_r^2}}} \right]\left[ {3{V_r} - 1} \right] = 8{T_r}:
View answer and explanation
Correct option: D
All of the above
Option D (All of the above) is correct because the given equation represents the law of corresponding states, which relates reduced pressure, volume, and temperature.
118 Medium
118 of 262
Analyzing +4 | -1
Assertion: Ice melts faster at high altitude.

Reason : At high altitude, atmospheric pressure is high.
View answer and explanation
Correct option: D
If both the Assertion and Reason are incorrect.
Option D (If both the Assertion and Reason are incorrect) is correct because ice melts slower at high altitude due to lower atmospheric pressure, which decreases the melting point.
119 Easy
119 of 262
Remembering +4 | -1
The ratio among most probable velocity, average velocity and root mean square velocity is given by
View answer and explanation
Correct option: D
2   : 8/π :  3\sqrt 2 \,\,\,:\,\sqrt {8/\pi } \,:\,\,\sqrt 3
Option D (2    8/π :  3\sqrt 2 \,\,\,\:\sqrt {8/\pi } \,:\,\,\sqrt 3) is correct because the ratio of most probable velocity, average velocity, and root mean square velocity is given by 2:8π:3\sqrt{2} : \sqrt{\frac{8}{\pi}} : \sqrt{3}.
120 Medium
120 of 262
Application +4 | -1
A gaseous mixture was prepared by taking equal mole of CO\mathrm{CO} and N2\mathrm{N}_2. If the total pressure of the mixture was found 1 atmosphere, the partial pressure of the nitrogen (N2)\left(\mathrm{N}_2\right) in the mixture is
View answer and explanation
Correct option: B
0.5 atm0.5 \mathrm{~atm}
Option B (0.5 atm0.5 \mathrm{~atm}) is correct because the partial pressure of nitrogen is half of the total pressure when equal moles of CO and N2 are present.
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