JEE Chemistry States Of Matter MCQs Set 26

Chemistry Objective Questions and Answers: States Of Matter

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Multiple Choice Questions

Question 1: The interaction energy of London forces is inversely proportional to the sixth power of the distance between two particles, but its magnitude depends on the
(a) charge of the interacting particles
(b) mass of the interacting particles
(c) polarisability of the interacting particles
(d) strength of permanent dipoles in the particles
Show Answer & Explanation

Answer: (c) polarisability of the interacting particles

Explanation:
1. London (dispersion) forces arise from temporary dipoles induced in neighbouring atoms.
2. Easily polarised (large, electron-rich) particles have stronger London forces.

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Question 2: London forces are always ___I___ and the interaction energy is inversely proportional to the ___II___ power of the distance between the particles. I and II are
(a) repulsive, sixth
(b) attractive, fourth
(c) attractive, sixth
(d) repulsive, fourth
Show Answer & Explanation

Answer: (c) attractive, sixth

Explanation:
1. London forces are always attractive, with \( E \propto 1/r^6 \).

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Question 3: Which of the following exhibits the weakest intermolecular forces?
(a) \( \ce{NH3} \)
(b) HCl
(c) He
(d) \( \ce{H2O} \)
Show Answer & Explanation

Answer: (c) He

Explanation:
1. Helium atoms are non-polar and have very few electrons, so only very weak London forces act between them.

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Question 4: 600 cc of a gas at a pressure of 750 mm Hg is compressed to 500 cc at constant temperature. The increase in pressure is
(a) 150 mm Hg
(b) 250 mm Hg
(c) 350 mm Hg
(d) 450 mm Hg
Show Answer & Explanation

Answer: (a) 150 mm Hg

Explanation:
1. \( P_2 = \frac{750 \times 600}{500} = 900 \) mm Hg.
2. Increase = 900 - 750 = 150 mm Hg.

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Question 5: 500 mL of nitrogen at 27 °C is cooled to -5 °C at the same pressure. The new volume is
(a) 326.32 mL
(b) 446.66 mL
(c) 546.32 mL
(d) 771.56 mL
Show Answer & Explanation

Answer: (b) 446.66 mL

Explanation:
1. \( V_2 = V_1 \frac{T_2}{T_1} = 500 \times \frac{268}{300} = 446.67 \) mL.

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Question 6: A balloon is filled with 2 L of air at 23.4 °C. The volume when the temperature rises to 26.1 °C at constant pressure is
(a) 2.018 L
(b) 2.8 L
(c) 3.5 L
(d) 1.5 L
Show Answer & Explanation

Answer: (a) 2.018 L

Explanation:
1. \( V_2 = 2 \times \frac{299.25}{296.55} = 2.018 \) L.

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Question 7: 4.4 g of a gas occupies 2.24 L at STP. The gas can be
(a) \( \ce{O2} \)
(b) CO
(c) \( \ce{NO2} \)
(d) \( \ce{CO2} \)
Show Answer & Explanation

Answer: (d) \( \ce{CO2} \)

Explanation:
1. 2.24 L is 0.1 mol, so the molar mass is 4.4/0.1 = 44 g/mol, which matches \( \ce{CO2} \).

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Question 8: Pure \( \ce{H2S} \) is stored in a 100 L tank at 20 °C and 2 atm. The mass of the gas is
(a) 34 g
(b) 340 g
(c) 282.68 g
(d) 28.24 g
Show Answer & Explanation

Answer: (c) 282.68 g

Explanation:
1. \( n = \frac{PV}{RT} = \frac{2 \times 100}{0.0821 \times 293} = 8.31 \) mol.
2. Mass = 8.31 x 34 = 282.7 g.

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Question 9: At NTP the volume of a gas is 273 mL. Its volume at 600 mm Hg and 273 °C is
(a) 391.8 mL
(b) 380 mL
(c) 691.6 mL
(d) 750 mL
Show Answer & Explanation

Answer: (c) 691.6 mL

Explanation:
1. \( V_2 = \frac{P_1 V_1 T_2}{T_1 P_2} = \frac{760 \times 273 \times 546}{273 \times 600} = 691.6 \) mL.

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Question 10: The pressure of a mixture of 4 g of \( \ce{O2} \) and 2 g of \( \ce{H2} \) confined in a 1 L bulb at 0 °C is
(a) 25.215 atm
(b) 31.205 atm
(c) 45.215 atm
(d) 15.210 atm
Show Answer & Explanation

Answer: (a) 25.215 atm

Explanation:
1. n = 4/32 + 2/2 = 1.125 mol.
2. \( P = \frac{nRT}{V} = 1.125 \times 0.0821 \times 273 / 1 = 25.2 \) atm.

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Question 11: The pressure exerted by 6.0 g of methane in a 0.03 m³ vessel at 129 °C is (C = 12.01, H = 1.01, R = 8.314 J/K/mol)
(a) 31684 Pa
(b) 215216 Pa
(c) 13409 Pa
(d) 41648 Pa
Show Answer & Explanation

Answer: (d) 41648 Pa

Explanation:
1. M = 16.05 g/mol, so n = 0.3738 mol.
2. \( P = \frac{0.3738 \times 8.314 \times 402}{0.03} = 4.16 \times 10^4 \) Pa.

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Question 12: An air bubble at 15 °C and 1.5 bar rises to the surface where the temperature is 25 °C and the pressure is 1.0 bar. The volume of the bubble
(a) becomes greater by a factor of 1.6
(b) becomes greater by a factor of 1.1
(c) becomes smaller by a factor of 0.70
(d) becomes greater by a factor of 2.5
Show Answer & Explanation

Answer: (a) becomes greater by a factor of 1.6

Explanation:
1. \( \frac{V_2}{V_1} = \frac{P_1 T_2}{P_2 T_1} = \frac{1.5 \times 298}{1.0 \times 288} = 1.55 \approx 1.6 \).

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Question 13: A mixture contains 64 g of dioxygen and 60 g of neon at a total pressure of 10 bar. The partial pressures of dioxygen and neon are respectively
(a) 4 and 6 bar
(b) 6 and 4 bar
(c) 5 and 5 bar
(d) 8 and 2 bar
Show Answer & Explanation

Answer: (a) 4 and 6 bar

Explanation:
1. Moles: \( \ce{O2} \) = 64/32 = 2, Ne = 60/20 = 3, so mole fractions 0.4 and 0.6.
2. Partial pressures = 4 bar and 6 bar.

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Question 14: 500 mL of air at 760 mm pressure is compressed to 200 mL at constant temperature. The pressure after compression is
(a) 1800 mm
(b) 1900 mm
(c) 2000 mm
(d) 1500 mm
Show Answer & Explanation

Answer: (b) 1900 mm

Explanation:
1. \( P_2 = \frac{760 \times 500}{200} = 1900 \) mm.

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Question 15: Two vessels of 16.4 L and 5 L contain ideal gases at 27 °C and 227 °C exerting 1.5 and 4.1 atm. The ratio of the number of molecules in the former to the latter is
(a) 2
(b) 1
(c) 1/2
(d) 1/3
Show Answer & Explanation

Answer: (a) 2

Explanation:
1. \( n_1 = \frac{1.5 \times 16.4}{R \times 300} \) and \( n_2 = \frac{4.1 \times 5}{R \times 500} \).
2. Ratio = 0.082 / 0.041 = 2.

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Question 16: 56 g of nitrogen and 96 g of oxygen are mixed at a total pressure of 10 atm. The partial pressures of oxygen and nitrogen in atm are respectively
(a) 4, 6
(b) 5, 5
(c) 2, 8
(d) 6, 4
Show Answer & Explanation

Answer: (d) 6, 4

Explanation:
1. \( n_{N_2} \) = 56/28 = 2 and \( n_{O_2} \) = 96/32 = 3, so the mole fractions are 0.4 and 0.6.
2. \( p_{O_2} \) = 6 atm and \( p_{N_2} \) = 4 atm.

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Question 17: If \( 10^{-4} \) dm³ of water is introduced into a 1.0 dm³ flask at 300 K, how many moles of water are in the vapour phase at equilibrium? (Vapour pressure of water at 300 K = 3170 Pa)
(a) \( 5.56 \times 10^{-3} \) mol
(b) \( 1.53 \times 10^{-2} \) mol
(c) \( 4.46 \times 10^{-2} \) mol
(d) \( 1.27 \times 10^{-3} \) mol
Show Answer & Explanation

Answer: (d) \( 1.27 \times 10^{-3} \) mol

Explanation:
1. \( n = \frac{PV}{RT} = \frac{3170 \times 1.0 \times 10^{-3}}{8.314 \times 300} = 1.27 \times 10^{-3} \) mol.
2. The 0.1 g of water (\( 5.56 x 10^{-3} \) mol) is more than this, so equilibrium is reached with some liquid left.

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Question 18: The deviation of a real gas from the ideal gas equation is more at
(a) high temperature and low pressure
(b) low temperature and high pressure
(c) high temperature and high pressure
(d) low temperature and low pressure
Show Answer & Explanation

Answer: (b) low temperature and high pressure

Explanation:
1. At low temperature the molecules move slowly and attract each other, and at high pressure the molecular volume is no longer negligible.

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Question 19: In van der Waals' equation, the constant 'b' is a measure of
(a) the volume occupied by the molecules
(b) intermolecular attraction
(c) intermolecular repulsion
(d) intermolecular collisions per unit volume
Show Answer & Explanation

Answer: (a) the volume occupied by the molecules

Explanation:
1. b is the excluded volume, approximately four times the actual molecular volume per mole.

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Question 20: The van der Waals constants 'a' for \( \ce{O2} \), \( \ce{N2} \), \( \ce{NH3} \) and \( \ce{CH4} \) are 1.360, 1.390, 4.170 and 2.253 L² atm mol⁻². The gas which can be most easily liquefied is
(a) \( \ce{O2} \)
(b) \( \ce{N2} \)
(c) \( \ce{NH3} \)
(d) \( \ce{CH4} \)
Show Answer & Explanation

Answer: (c) \( \ce{NH3} \)

Explanation:
1. A larger value of a means stronger intermolecular attraction and easier liquefaction, so \( \ce{NH3} \) liquefies most easily.

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Question 21: The unit of the van der Waals constant 'a' is
(a) \( \ce{dm^6 atm mol^{-2}} \)
(b) \( \ce{dm^3 atm mol^{-1}} \)
(c) \( \ce{dm atm mol^{-1}} \)
(d) \( \ce{atm mol^{-1}} \)
Show Answer & Explanation

Answer: (a) \( \ce{dm^6 atm mol^{-2}} \)

Explanation:
1. The term \( a/V^2 \) must have the units of pressure, so a has units of atm L² mol⁻² = atm dm⁶ mol⁻².

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Question 22: The van der Waals constant 'a' for gases P, Q, R and S are 4.17, 3.59, 6.71 and 3.8 atm L² mol⁻². The ascending order of their liquefaction (easiest last) is
(a) R < P < S < Q
(b) Q < S < R < P
(c) Q < S < P < R
(d) R < P < Q < S
Show Answer & Explanation

Answer: (c) Q < S < P < R

Explanation:
1. The larger the value of a, the easier the gas is to liquefy, so the order of ease is Q (3.59) < S (3.8) < P (4.17) < R (6.71).

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Question 23: The gas with the highest critical temperature among those listed is
(a) \( \ce{H2} \)
(b) He
(c) \( \ce{N2} \)
(d) \( \ce{CO2} \)
Show Answer & Explanation

Answer: (d) \( \ce{CO2} \)

Explanation:
1. The critical temperatures are H2 33 K, He 5.3 K, N2 126 K and CO2 304 K, so \( \ce{CO2} \) has the highest.

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Question 24: A real gas behaves like an ideal gas when
(a) the temperature is low
(b) both the temperature and the pressure are low
(c) both the temperature and the pressure are high
(d) the temperature is high and the pressure is low
Show Answer & Explanation

Answer: (d) the temperature is high and the pressure is low

Explanation:
1. At high temperature and low pressure the molecular volume and attractions are negligible.

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Question 25: An ideal gas cannot be liquefied because
(a) its critical temperature is always above 0 °C
(b) its molecules are relatively small
(c) it solidifies before becoming a liquid
(d) forces between its molecules are negligible
Show Answer & Explanation

Answer: (d) forces between its molecules are negligible

Explanation:
1. Liquefaction needs intermolecular attraction; an ideal gas has none.

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

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