JEE Chemistry States Of Matter MCQs Set 19

Chemistry Objective Questions and Answers: States Of Matter

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Multiple Choice Questions

Question 1: The increasing order of effusion rate among the gases H2, O2, NH3 and CO2 is:
(a) H2, CO2, NH3, O2
(b) H2, NH3, O2, CO2
(c) H2, O2, NH3, CO2
(d) CO2, O2, NH3, H2
Show Answer & Explanation

Answer: (d) CO2, O2, NH3, H2

Explanation:
1. Graham's law: rate of effusion or diffusion is proportional to \( \frac{1}{\sqrt{M}} \) at the same T and P.
2. Molar masses: H2 = 2, NH3 = 17, O2 = 32, CO2 = 44.
3. The heavier the gas, the slower it effuses, so the increasing order of rate is CO2 < O2 < NH3 < H2.
4. Answer: (d)

Teacher's Note:
1. Rank gases by molar mass first.
2. Increasing rate means decreasing molar mass.

Question 2: A gas A of molecular mass 4 diffuses three times as fast as a gas B at the same temperature and pressure. The molecular mass of B is:
(a) 36
(b) 12
(c) 18
(d) 24
Show Answer & Explanation

Answer: (a) 36

Explanation:
1. Graham's law: rate of effusion or diffusion is proportional to \( \frac{1}{\sqrt{M}} \) at the same T and P.
2. \( \frac{r_A}{r_B}=\sqrt{\frac{M_B}{M_A}}=3 \)
3. \( M_B = 9 \times 4 = 36 \)
4. Answer: (a)

Teacher's Note:
1. Square the rate ratio to get the mass ratio.
2. The gas that is three times faster is nine times lighter.

Question 3: Four rubber tubes are filled with H2, O2, N2 and CO2 respectively, each at the same pressure. Through tiny pores in the rubber, the tube that deflates first is the one filled with:
(a) H2
(b) O2
(c) N2
(d) CO2
Show Answer & Explanation

Answer: (a) H2

Explanation:
1. Graham's law: rate of effusion or diffusion is proportional to \( \frac{1}{\sqrt{M}} \) at the same T and P.
2. Molar masses: H2 = 2, O2 = 32, N2 = 28, CO2 = 44.
3. H2 has the smallest molar mass, so it escapes fastest and its tube deflates first.
4. Answer: (a)

Teacher's Note:
1. Lightest gas leaks out fastest.
2. Compare molar masses, not formula size.

Question 4: A balloon filled with methane (CH4) is pricked with a sharp point and quickly plunged into a tank of hydrogen at the same pressure. After some time the balloon will:
(a) Be enlarged
(b) Collapse
(c) Remain unchanged in size
(d) Have ethylene (C2H4) inside it
Show Answer & Explanation

Answer: (a) Be enlarged

Explanation:
1. Graham's law: rate of effusion or diffusion is proportional to \( \frac{1}{\sqrt{M}} \) at the same T and P.
2. H2 (M = 2) diffuses into the balloon faster than CH4 (M = 16) diffuses out.
3. More gas enters than leaves, so the balloon swells.
4. Answer: (a)

Teacher's Note:
1. Compare the two diffusion rates.
2. The net flow is towards the side of the slower gas.

Question 5: A football bladder contains equimolar H2 and O2. The composition by mass of the mixture effusing out of a puncture, H2 : O2, is:
(a) 1 : 4
(b) \( 2\sqrt{2} : 1 \)
(c) \( 1 : 2\sqrt{2} \)
(d) 4 : 1
Show Answer & Explanation

Answer: (a) 1 : 4

Explanation:
1. Graham's law: rate of effusion or diffusion is proportional to \( \frac{1}{\sqrt{M}} \) at the same T and P.
2. Mole ratio effusing = rate ratio = \( \sqrt{32/2}=4 \), so H2 : O2 = 4 : 1 by moles.
3. Mass ratio = (4 × 2) : (1 × 32) = 8 : 32 = 1 : 4.
4. Answer: (a)

Teacher's Note:
1. Equimolar mixture: rate ratio equals mole ratio of the effusate.
2. Convert moles to mass with the molar masses.

Question 6: A gas X diffuses three times faster than another gas Y. The ratio of their densities \( d_X : d_Y \) is:
(a) 1/3
(b) 1/9
(c) 1/6
(d) 1/12
Show Answer & Explanation

Answer: (b) 1/9

Explanation:
1. Graham's law: rate of effusion or diffusion is proportional to \( \frac{1}{\sqrt{M}} \) at the same T and P. (equivalently, proportional to \( 1/\sqrt{d} \)).
2. \( \frac{r_X}{r_Y}=\sqrt{\frac{d_Y}{d_X}}=3 \Rightarrow \frac{d_X}{d_Y}=\frac{1}{9} \)
3. Answer: (b)

Teacher's Note:
1. Square the rate ratio.
2. Faster gas means lower density.

Question 7: The relative rate of diffusion of a gas of molecular mass 128 compared with oxygen is:
(a) 2 times
(b) 1/4
(c) 1/8
(d) 1/2
Show Answer & Explanation

Answer: (d) 1/2

Explanation:
1. Graham's law: rate of effusion or diffusion is proportional to \( \frac{1}{\sqrt{M}} \) at the same T and P.
2. \( \frac{r}{r_{O_2}}=\sqrt{\frac{32}{128}}=\sqrt{\frac14}=\frac12 \)
3. Answer: (d)

Teacher's Note:
1. Put the lighter molar mass in the numerator.
2. 128 is four times 32.

Question 8: The atomic masses of C, N and O are 12, 14 and 16. Among the following pairs of gases, the pair that diffuses at the same rate is:
(a) Carbon dioxide and nitrous oxide
(b) Carbon dioxide and nitrogen peroxide
(c) Carbon dioxide and carbon monoxide
(d) Carbon dioxide and nitric oxide
Show Answer & Explanation

Answer: (a) Carbon dioxide and nitrous oxide

Explanation:
1. Same rate requires same molar mass.
2. CO2 = 44, N2O = 28 + 16 = 44, NO2 = 46, CO = 28, NO = 30.
3. Only CO2 and N2O match.
4. Answer: (a)

Teacher's Note:
1. Calculate each molar mass.
2. Equal molar masses imply equal rates.

Question 9: A bottle of dry ammonia and a bottle of dry hydrogen chloride are connected through a long tube and opened simultaneously at both ends. The white ammonium chloride ring first formed will be:
(a) At the centre of the tube
(b) Near the hydrogen chloride bottle
(c) Near the ammonia bottle
(d) Throughout the length of the tube
Show Answer & Explanation

Answer: (b) Near the hydrogen chloride bottle

Explanation:
1. NH3 (M = 17) diffuses faster than HCl (M = 36.5).
2. NH3 travels farther before meeting HCl, so the ring forms closer to the HCl end.
3. Answer: (b)

Teacher's Note:
1. The faster gas covers more distance.
2. The ring forms where the two gases meet.

Question 10: 50 mL of a gas A diffuses through a membrane in the same time as 40 mL of a gas B under identical conditions. If the molecular mass of A is 64, that of B is:
(a) 100
(b) 250
(c) 200
(d) 80
Show Answer & Explanation

Answer: (a) 100

Explanation:
1. Graham's law: rate of effusion or diffusion is proportional to \( \frac{1}{\sqrt{M}} \) at the same T and P.
2. \( \frac{r_A}{r_B}=\frac{50}{40}=\sqrt{\frac{M_B}{64}} \)
3. \( M_B = 64 \times \frac{25}{16}=100 \)
4. Answer: (a)

Teacher's Note:
1. Same time means rate ratio equals volume ratio.
2. Square 5/4 and multiply by 64.

Question 11: 50 mL of hydrogen diffuses through a small hole from a vessel in 20 minutes. The time taken by 40 mL of oxygen to diffuse out under similar conditions is:
(a) 12 min
(b) 64 min
(c) 8 min
(d) 32 min
Show Answer & Explanation

Answer: (b) 64 min

Explanation:
1. Rate of H2 = 50/20 = 2.5 mL/min.
2. \( r_{O_2}=2.5\sqrt{2/32}=0.625 \) mL/min.
3. Time = 40/0.625 = 64 min.
4. Answer: (b)

Teacher's Note:
1. Find the rate of H₂ first.
2. Scale it by the root of the molar mass ratio.

Question 12: The rate of diffusion of a gas whose molecular mass is exactly double that of nitrogen is 56 mL per second. The rate of diffusion of nitrogen under the same conditions is:
(a) 79.19 mL/s
(b) 112 mL/s
(c) 56 mL/s
(d) 90 mL/s
Show Answer & Explanation

Answer: (a) 79.19 mL/s

Explanation:
1. Graham's law: rate of effusion or diffusion is proportional to \( \frac{1}{\sqrt{M}} \) at the same T and P.
2. \( \frac{r_{N_2}}{56}=\sqrt{2} \)
3. \( r_{N_2}=56\sqrt2 \approx 79.19 \) mL/s
4. Answer: (a)

Teacher's Note:
1. Nitrogen is the lighter gas, so it is faster.
2. Multiply by the root of 2.

Question 13: Under identical conditions of temperature and pressure, the ratio of the rates of effusion of O2 and CO2 is:
(a) 0.87
(b) 1.17
(c) 8.7
(d) 0.117
Show Answer & Explanation

Answer: (b) 1.17

Explanation:
1. Graham's law: rate of effusion or diffusion is proportional to \( \frac{1}{\sqrt{M}} \) at the same T and P.
2. \( \frac{r_{O_2}}{r_{CO_2}}=\sqrt{\frac{44}{32}}=\sqrt{1.375}=1.17 \)
3. Answer: (b)

Teacher's Note:
1. Lighter gas on the top of the ratio gives a value above 1.
2. Use molar masses 32 and 44.

Question 14: A real gas shows behaviour the same as an ideal gas at:
(a) Low temperature and low pressure
(b) High temperature and high pressure
(c) Low temperature and high pressure
(d) High temperature and low pressure
Show Answer & Explanation

Answer: (d) High temperature and low pressure

Explanation:
1. At low pressure the molecules are far apart, so their volume and mutual attraction are negligible.
2. At high temperature the kinetic energy overcomes the attractions.
3. So the ideal gas law holds at high temperature and low pressure.
4. Answer: (d)

Teacher's Note:
1. Think of the two assumptions of kinetic theory.
2. Both are satisfied at high T and low P.

Question 15: In van der Waals' equation of state, the constant 'b' is a measure of:
(a) Intermolecular repulsions
(b) Intermolecular attraction
(c) Volume occupied by the molecules
(d) Intermolecular collisions per unit volume
Show Answer & Explanation

Answer: (c) Volume occupied by the molecules

Explanation:
1. The volume correction is V − nb.
2. b equals four times the actual volume of one mole of molecules (excluded volume).
3. Answer: (c)

Teacher's Note:
1. b corrects the volume term.
2. a corrects the pressure term.

Question 16: The term that accounts for intermolecular forces in van der Waals' equation for a non-ideal gas is:
(a) RT
(b) V − b
(c) \( \left(P+\frac{a}{V^2}\right) \)
(d) [RT]−1
Show Answer & Explanation

Answer: (c) \( \left(P+\frac{a}{V^2}\right) \)

Explanation:
1. The attraction between molecules lowers the pressure exerted on the walls.
2. The correction \( \frac{a}{V^2} \) is added to P to restore the ideal pressure.
3. Answer: (c)

Teacher's Note:
1. The pressure term carries the effect of attraction.
2. The volume term carries the effect of size.

Question 17: The pressure of a real gas is less than that of an ideal gas because of:
(a) Increase in the number of collisions
(b) Finite size of the molecules
(c) Increase in the kinetic energy of the molecules
(d) Intermolecular forces
Show Answer & Explanation

Answer: (d) Intermolecular forces

Explanation:
1. Attractive forces pull a molecule near the wall inward.
2. The impact on the wall is reduced, so the measured pressure is lower than the ideal value.
3. Answer: (d)

Teacher's Note:
1. Finite size affects volume, not pressure directly.
2. Attraction reduces wall impacts.

Question 18: Given the van der Waals constant 'a' (atm L2 mol−2): NH3 = 4.17, CO2 = 3.59, SO2 = 6.71, Cl2 = 6.49. Which gas can be most easily liquefied?
(a) NH3
(b) Cl2
(c) SO2
(d) CO2
Show Answer & Explanation

Answer: (c) SO2

Explanation:
1. A larger value of a means stronger intermolecular attraction.
2. Stronger attraction makes liquefaction easier.
3. SO2 has the largest a (6.71).
4. Answer: (c)

Teacher's Note:
1. Compare the a values.
2. Highest a means easiest liquefaction.

Question 19: At relatively high pressure, the van der Waals equation (for one mole) reduces to:
(a) PV = RT
(b) \( PV = RT + \frac{a}{V} \)
(c) PV = RT + Pb
(d) \( PV = RT - \frac{a}{V^2} \)
Show Answer & Explanation

Answer: (c) PV = RT + Pb

Explanation:
1. \( \left(P+\frac{a}{V^2}\right)(V-b)=RT \)
2. At high pressure the molar volume V is small but P is very large, so a/V² is negligible compared with P, while b is not negligible compared with V.
3. P(V − b) = RT, i.e. PV = RT + Pb.
4. Answer: (c)

Teacher's Note:
1. Decide which correction survives.
2. At high pressure, the excluded volume dominates.

Question 20: A real gas most closely approaches the behaviour of an ideal gas at:
(a) 15 atm and 200 K
(b) 1 atm and 273 K
(c) 0.5 atm and 500 K
(d) 15 atm and 500 K
Show Answer & Explanation

Answer: (c) 0.5 atm and 500 K

Explanation:
1. Ideal behaviour needs the lowest pressure and the highest temperature.
2. Among the options, 0.5 atm and 500 K has both.
3. Answer: (c)

Teacher's Note:
1. Look for low P and high T together.
2. Check both conditions, not just one.

Question 21: The compressibility factor of an ideal gas is:
(a) 0
(b) 1
(c) 2
(d) 4
Show Answer & Explanation

Answer: (b) 1

Explanation:
1. Z = PV/nRT.
2. For an ideal gas PV = nRT, hence Z = 1.
3. Answer: (b)

Teacher's Note:
1. Use the definition of Z.
2. Z = 1 at all P and T for an ideal gas.

Question 22: The compressibility factor of a gas is less than unity at STP. Therefore:
(a) Vm > 22.4 L
(b) Vm < 22.4 L
(c) Vm = 22.4 L
(d) Vm = 44.8 L
Show Answer & Explanation

Answer: (b) Vm < 22.4 L

Explanation:
1. \( Z=\frac{V_m}{V_{ideal}}=\frac{V_m}{22.4}<1 \)
2. So Vm < 22.4 L.
3. Answer: (b)

Teacher's Note:
1. Ideal molar volume at STP is 22.4 L.
2. Z is the ratio of real to ideal volume.

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