JEE Chemistry States Of Matter MCQs Set 06

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Multiple Choice Questions

Question 1: A 1 L flask contains \( \ce{N2} \) at 100 kPa and a 3 L flask contains \( \ce{O2} \) at 320 kPa. If the two flasks are connected, the resultant pressure is
(a) 310 kPa
(b) 210 kPa
(c) 420 kPa
(d) 265 kPa
Show Answer & Explanation

Answer: (d) 265 kPa

Explanation:
1. Each gas expands into the total volume of 4 L at constant temperature (Boyle's law).
2. Total \( PV=100\times1+320\times3=1060 \) kPa L.
3. P = 1060/4 = 265 kPa.

Teacher's Note:
1. Gases do not react here, so partial pressures simply add.
2. Use \( P_1V_1+P_2V_2=P_{total}V_{total} \).

Question 2: 10 g of a gas at atmospheric pressure is cooled from \( 273^\circ C \) to \( 0^\circ C \) keeping the volume constant. Its pressure would become
(a) 1/2 atm
(b) 1/273 atm
(c) 2 atm
(d) 273 atm
Show Answer & Explanation

Answer: (a) 1/2 atm

Explanation:
1. Gay-Lussac's law at constant V: \( P_1/T_1=P_2/T_2 \).
2. T1 = 546 K, T2 = 273 K.
3. P2 = 1 \( \times \) 273/546 = 1/2 atm.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 3: A gas sample has volume 0.2 L at 1 atm and \( 0^\circ C \). At the same pressure but at \( 273^\circ C \) its volume will be
(a) 0.4 litres
(b) 0.8 litres
(c) 27.8 litres
(d) 55.6 litres
Show Answer & Explanation

Answer: (a) 0.4 litres

Explanation:
1. Charles' law: \( V_1/T_1=V_2/T_2 \).
2. V2 = 0.2 \( \times \) 546/273 = 0.4 L.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 4: 400 \( cm^3 \) of oxygen at \( 27^\circ C \) were cooled to \( -3^\circ C \) without change in pressure. The contraction in volume is
(a) \( 40\ cm^3 \)
(b) \( 30\ cm^3 \)
(c) \( 44.4\ cm^3 \)
(d) \( 360\ cm^3 \)
Show Answer & Explanation

Answer: (a) \( 40\ cm^3 \)

Explanation:
1. T1 = 300 K, T2 = 270 K.
2. V2 = 400 \( \times \) 270/300 = 360 \( cm^3 \).
3. Contraction = 400 - 360 = 40 \( cm^3 \).

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 5: 1.0 g of \( \ce{H2} \) in a closed 5 L flask is heated from 300 K to 600 K. Which statement is NOT correct?
(a) Pressure of the gas increases
(b) The rate of collisions increases
(c) The number of moles of gas increases
(d) The energy of the gas molecules increases
Show Answer & Explanation

Answer: (c) The number of moles of gas increases

Explanation:
1. The flask is closed, so the amount of gas is fixed at 0.5 mol.
2. Heating raises the speed, collision rate, kinetic energy and (at constant V) the pressure.
3. The number of moles does not change, so (c) is incorrect.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 6: Two closed vessels of equal volume containing air at pressure \( P_1 \) and temperature \( T_1 \) are connected by a narrow tube. One vessel is kept at \( T_1 \) and the other at \( T_2 \). The pressure in the vessels will be
(a) \( \frac{2P_1T_1}{T_1+T_2} \)
(b) \( \frac{T_1}{2P_1T_2} \)
(c) \( \frac{2P_1T_2}{T_1+T_2} \)
(d) \( \frac{2P_1}{T_1+T_2} \)
Show Answer & Explanation

Answer: (c) \( \frac{2P_1T_2}{T_1+T_2} \)

Explanation:
1. Total moles stay constant: \( \frac{2P_1V}{RT_1}=\frac{PV}{RT_1}+\frac{PV}{RT_2} \).
2. Hence \( \frac{2P_1}{T_1}=P\left(\frac1{T_1}+\frac1{T_2}\right) \).
3. P = \( \frac{2P_1T_2}{T_1+T_2} \).

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 7: 16 g of oxygen and 3 g of hydrogen are mixed and kept at 760 mm pressure and \( 0^\circ C \). The total volume occupied by the mixture is nearly
(a) 22.4 litres
(b) 33.6 litres
(c) 448 litres
(d) 44800 mL
Show Answer & Explanation

Answer: (d) 44800 mL

Explanation:
1. n(\( \ce{O2} \)) = 16/32 = 0.5 mol; n(\( \ce{H2} \)) = 3/2 = 1.5 mol.
2. Total = 2 mol at STP.
3. V = 2 \( \times \) 22.4 = 44.8 L = 44800 mL.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 8: Pure hydrogen sulphide is stored in a 100 L tank at \( 20^\circ C \) and 2 atm. The mass of the gas is
(a) 34 g
(b) 340 g
(c) 282.4 g
(d) 28.24 g
Show Answer & Explanation

Answer: (c) 282.4 g

Explanation:
1. n = \( \frac{PV}{RT}=\frac{2\times100}{0.0821\times293} \) = 8.31 mol.
2. Mass = 8.31 \( \times \) 34 = 282.6 g, i.e. about 282.4 g.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

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

Answer: (c) 691.6 mL

Explanation:
1. \( \frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} \) with P1 = 760, T1 = 273 K, P2 = 600, T2 = 546 K.
2. V2 = \( 273\times\frac{760}{600}\times\frac{546}{273} \) = 691.6 mL.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 10: One litre of a gas weighs 2 g at 300 K and 1 atm. At 0.75 atm, at which temperature will one litre of the same gas weigh 1 g?
(a) 450 K
(b) 600 K
(c) 800 K
(d) 900 K
Show Answer & Explanation

Answer: (a) 450 K

Explanation:
1. Density \( d=\frac{PM}{RT} \), so \( \frac{d_1}{d_2}=\frac{P_1T_2}{P_2T_1} \).
2. 2/1 = \( \frac{1\times T_2}{0.75\times300} \).
3. T2 = 2 \( \times \) 225 = 450 K.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 11: A weather balloon filled with hydrogen at 1 atm and \( 27^\circ C \) has volume 12000 L. On ascending it reaches a place where the temperature is \( -23^\circ C \) and pressure is 0.5 atm. The volume of the balloon is
(a) 24000 litres
(b) 20000 litres
(c) 10000 litres
(d) 12000 litres
Show Answer & Explanation

Answer: (b) 20000 litres

Explanation:
1. \( V_2=V_1\frac{P_1T_2}{P_2T_1} \).
2. V2 = 12000 \( \times\frac{1\times250}{0.5\times300} \) = 20000 L.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 12: The density of a gas at \( 27^\circ C \) and 1 atm is d. Pressure remaining constant, at which temperature will its density become 0.75 d?
(a) \( 20^\circ C \)
(b) \( 30^\circ C \)
(c) 400 K
(d) 300 K
Show Answer & Explanation

Answer: (c) 400 K

Explanation:
1. At constant P, density is inversely proportional to T.
2. T2 = 300/0.75 = 400 K.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 13: A gas occupies 100 mL at \( 27^\circ C \) and 740 mm pressure. When its volume is changed to 80 mL at the same pressure, the temperature will be
(a) \( 21.6^\circ C \)
(b) \( 240^\circ C \)
(c) \( -33^\circ C \)
(d) \( 89.5^\circ C \)
Show Answer & Explanation

Answer: (c) \( -33^\circ C \)

Explanation:
1. Charles' law: \( V_1/T_1=V_2/T_2 \).
2. T2 = 300 \( \times \) 80/100 = 240 K.
3. 240 - 273 = \( -33^\circ C \).

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 14: If two moles of an ideal gas at 546 K occupy a volume of 44.8 L, the pressure must be
(a) 2 atm
(b) 3 atm
(c) 4 atm
(d) 1 atm
Show Answer & Explanation

Answer: (a) 2 atm

Explanation:
1. P = \( \frac{nRT}{V}=\frac{2\times0.0821\times546}{44.8} \) = 2.0 atm.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 15: At \( 0^\circ C \) and 1 atm a gas occupies 100 cc. If the pressure is increased to one and a half times and the absolute temperature is increased by one-third, the final volume is
(a) 80 cc
(b) 88.9 cc
(c) 66.7 cc
(d) 100 cc
Show Answer & Explanation

Answer: (b) 88.9 cc

Explanation:
1. P2 = 1.5 atm; T2 = 273 \( \times \) 4/3 = 364 K.
2. V2 = \( 100\times\frac{1}{1.5}\times\frac{4}{3} \) = 88.9 cc.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 16: Hydrogen diffuses six times faster than gas A. The molar mass of A is
(a) 72
(b) 6
(c) 24
(d) 36
Show Answer & Explanation

Answer: (a) 72

Explanation:
1. Graham's law: \( \frac{r_H}{r_A}=\sqrt{\frac{M_A}{M_H}} \).
2. 6 = \( \sqrt{M_A/2} \) so \( M_A=36\times2=72 \).

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 17: 4.4 g of a gas at STP occupies 2.24 L. 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 at STP = 0.1 mol.
2. M = 4.4/0.1 = 44 g/mol.
3. Only \( \ce{CO2} \) has a molar mass of 44 g/mol.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 18: \( 6.02\times10^{22} \) molecules each of \( \ce{N2} \), \( \ce{O2} \) and \( \ce{H2} \) are mixed. The mass of the mixture in grams is
(a) 6.2
(b) 4.12
(c) 3.09
(d) 7
Show Answer & Explanation

Answer: (a) 6.2

Explanation:
1. Each gas is 0.1 mol.
2. Mass = 0.1(28 + 32 + 2) = 6.2 g.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 19: A 5 L cylinder filled with air at NTP is connected to an evacuated 30 L cylinder. The resultant air pressure in both cylinders will be
(a) 10.8 cm of Hg
(b) 14.9 cm of Hg
(c) 21.8 cm of Hg
(d) 38.8 cm of Hg
Show Answer & Explanation

Answer: (a) 10.8 cm of Hg

Explanation:
1. Initial pressure at NTP = 76 cm Hg.
2. Boyle's law: P2 = 76 \( \times \) 5/35 = 10.86 cm Hg.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 20: A certain mass of gas occupies 300 c.c. at \( 27^\circ C \) and 620 mm pressure. Its volume at \( 47^\circ C \) and 640 mm pressure will be
(a) 400 c.c.
(b) 510 c.c.
(c) 310 c.c.
(d) 350 c.c.
Show Answer & Explanation

Answer: (c) 310 c.c.

Explanation:
1. V2 = \( 300\times\frac{620}{640}\times\frac{320}{300} \) = 310 c.c.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 21: What will be the volume of the mixture after the reaction \( \ce{NH3 + HCl -> NH4Cl} \) if 4 L of \( \ce{NH3} \) is mixed with 1.5 L of HCl (both at the same conditions; \( \ce{NH4Cl} \) is a solid)?
(a) 0.5 litre
(b) 1 litre
(c) 2.5 litres
(d) 0.1 litre
Show Answer & Explanation

Answer: (c) 2.5 litres

Explanation:
1. The gases react in the volume ratio 1 : 1.
2. 1.5 L of HCl consumes 1.5 L of \( \ce{NH3} \), forming solid \( \ce{NH4Cl} \).
3. Remaining \( \ce{NH3} \) = 4 - 1.5 = 2.5 L.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 22: The ratio of the root mean square velocity to the average velocity of gas molecules at a particular temperature is
(a) 1.086 : 1
(b) 1 : 1.086
(c) 2 : 1.086
(d) 1.086 : 2
Show Answer & Explanation

Answer: (a) 1.086 : 1

Explanation:
1. \( u_{rms}=\sqrt{3RT/M} \) and \( u_{av}=\sqrt{8RT/\pi M} \).
2. Ratio = \( \sqrt{3\pi/8} \) = 1.086 : 1.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 23: The ratio \( C_p/C_v \) of a gas X is 1.4. The number of atoms of X present in 11.2 L of it at NTP is
(a) \( 6.02\times10^{23} \)
(b) \( 1.2\times10^{24} \)
(c) \( 3.01\times10^{23} \)
(d) \( 2.01\times10^{23} \)
Show Answer & Explanation

Answer: (a) \( 6.02\times10^{23} \)

Explanation:
1. A ratio of 1.4 indicates a diatomic gas.
2. 11.2 L at NTP = 0.5 mol of molecules = 1 mol of atoms.
3. Number of atoms = \( 6.02\times10^{23} \).

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 24: The density of neon will be highest at
(a) S.T.P.
(b) \( 0^\circ C,\ 2\ atm \)
(c) \( 273^\circ C,\ 1\ atm \)
(d) \( 273^\circ C,\ 2\ atm \)
Show Answer & Explanation

Answer: (b) \( 0^\circ C,\ 2\ atm \)

Explanation:
1. Density is proportional to P/T.
2. P/T values: 1/273, 2/273, 1/546 and 2/546.
3. The largest is 2/273, i.e. 0 \( ^\circ C \) and 2 atm.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 25: If the inversion temperature of a gas is \( -80^\circ C \), it will produce cooling under the Joule-Thomson effect at
(a) 298 K
(b) 273 K
(c) 193 K
(d) 173 K
Show Answer & Explanation

Answer: (d) 173 K

Explanation:
1. Inversion temperature = 273 - 80 = 193 K.
2. A gas cools on expansion only below its inversion temperature.
3. Of the options only 173 K is below 193 K.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

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