JEE Chemistry States of Matter MCQs Set 03

Answer: (d) NO, C₂H₆

Explanation:
1. At the same P and T the rate of diffusion depends only on molar mass: \( r\propto\frac{1}{\sqrt{M}} \).
2. Molar masses: NO = 30, CO = 28, CO₂ = 44, NH₃ = 17, PH₃ = 34, C₂H₆ = 30.
3. Only NO and C₂H₆ have equal molar masses (30), so they diffuse at the same rate.

Teacher's Note:
1. Equal rates mean equal molar masses.
2. Add up atomic masses carefully for each pair.

Answer: (a) 64

Explanation:
1. \( \frac{r_{CH_4}}{r_X}=\sqrt{\frac{M_X}{M_{CH_4}}}=2 \).
2. So \( M_X=4\times16=64 \).

Teacher's Note:
1. Square both sides to remove the root.
2. Methane has molar mass 16.

Answer: (d) CO₂, O₂, NH₃, H₂

Explanation:
1. Rate of effusion is inversely proportional to \( \sqrt M \).
2. Molar masses: CO₂ 44, O₂ 32, NH₃ 17, H₂ 2.
3. Increasing rate means decreasing molar mass: CO₂ < O₂ < NH₃ < H₂.

Teacher's Note:
1. Heaviest gas is slowest.
2. Read the order asked: increasing rate.

Answer: (a) 36

Explanation:
1. \( \frac{r_A}{r_B}=3=\sqrt{\frac{M_B}{M_A}} \).
2. So \( M_B=9\times4=36 \).

Teacher's Note:
1. Heavier gas is slower, so B must be heavier than A.
2. Square the ratio of rates.

Answer: (a) H₂ filled tube

Explanation:
1. Gas escapes through the pores by effusion, whose rate is proportional to \( 1/\sqrt M \).
2. H₂ has the smallest molar mass (2), so it escapes fastest and its tube deflates first.

Teacher's Note:
1. Lightest gas leaks fastest.
2. The source wording said re-inflated; rewritten as deflated, which is the intended physics.

Answer: (a) enlarged

Explanation:
1. Rate of diffusion is inversely proportional to \( \sqrt M \).
2. H₂ (M = 2) enters the balloon faster than CH₄ (M = 16) leaves it.
3. More gas enters than leaves, so the balloon enlarges.

Teacher's Note:
1. Compare the two rates, one inward and one outward.
2. The lighter gas wins.

Answer: (b) 1.4 times that of He

Explanation:
1. \( \frac{r_{H_2}}{r_{He}}=\sqrt{\frac{4}{2}}=\sqrt2\approx1.4 \).

Teacher's Note:
1. Use molar masses 2 and 4.
2. The ratio is the square root of 2.

Answer: (a) 1 : 4

Explanation:
1. For equal moles, the moles effusing are in the ratio of the rates: \( \frac{n_{H_2}}{n_{O_2}}=\sqrt{\frac{32}{2}}=4 \).
2. Mass ratio \( =\frac{4\times2}{1\times32}=\frac{8}{32}=\frac14 \).

Teacher's Note:
1. Convert from the mole ratio to the mass ratio at the end.
2. Do not stop at the rate ratio.

Answer: (b) 1 : 1.414

Explanation:
1. \( \frac{r_{O_2}}{r_{CH_4}}=\sqrt{\frac{d_{CH_4}}{d_{O_2}}}=\sqrt{\frac12}=\frac{1}{1.414} \).
2. So the ratio is 1 : 1.414.

Teacher's Note:
1. Denser gas diffuses more slowly.
2. Rate ratio is the inverse square root of the density ratio.

Answer: (b) 1/9

Explanation:
1. \( \frac{r_X}{r_Y}=\sqrt{\frac{D_Y}{D_X}}=3 \), so \( \frac{D_Y}{D_X}=9 \).
2. Hence \( D_X:D_Y=1:9 \).

Teacher's Note:
1. Invert carefully: the faster gas has the smaller density.
2. Square the rate ratio.

Answer: (a) 1/7

Explanation:
1. \( \frac{r_{gas}}{r_{H_2}}=\sqrt{\frac{2}{98}}=\frac17 \).

Teacher's Note:
1. Hydrogen has molar mass 2.
2. 98/2 = 49, whose root is 7.

Answer: (d) 1/2 times

Explanation:
1. \( \frac{r_{gas}}{r_{O_2}}=\sqrt{\frac{32}{128}}=\sqrt{\frac14}=\frac12 \).

Teacher's Note:
1. Oxygen has molar mass 32.
2. Heavier gas is slower, so the answer must be below 1.

Answer: (a) carbon dioxide and nitrous oxide

Explanation:
1. Equal rates need equal molar masses.
2. CO₂ = 44 and N₂O = 28 + 16 = 44.
3. Others: NO₂ = 46, CO = 28, NO = 30.

Teacher's Note:
1. Compute all molar masses.
2. N₂O and CO₂ are the classic equal-mass pair.

Answer: (b) near the hydrogen chloride bottle

Explanation:
1. NH₃ (M = 17) diffuses faster than HCl (M = 36.5).
2. The gases meet at a point nearer to the slower gas, so the ring forms near the HCl bottle.

Teacher's Note:
1. The faster gas covers more distance before they meet.
2. The ring is closer to the heavier gas.

Answer: (a) 100

Explanation:
1. \( \frac{r_A}{r_B}=\frac{50}{40}=\sqrt{\frac{M_B}{64}} \).
2. So \( M_B=64\times\frac{25}{16}=100 \).

Teacher's Note:
1. Equal time means the ratio of volumes is the ratio of rates.
2. Square 5/4 to get 25/16.

Answer: (a) 1/25

Explanation:
1. \( \frac{r_A}{r_B}=\sqrt{\frac{d_B}{d_A}}=5 \), so \( \frac{d_A}{d_B}=\frac{1}{25} \).

Teacher's Note:
1. The faster gas has the smaller density.
2. Square the rate ratio and invert.

Answer: (b) 64 min

Explanation:
1. Rate of H₂ = 50/20 = 2.5 mL/min.
2. Rate of O₂ = \( 2.5\times\sqrt{2/32}=2.5\times\frac14=0.625 \) mL/min.
3. Time = 40/0.625 = 64 min.

Teacher's Note:
1. Find the rate of hydrogen first.
2. Use the ratio 1:4 for O₂ to H₂.

Answer: (b) 4 : 1

Explanation:
1. \( \frac{r_1}{r_2}=\sqrt{\frac{d_2}{d_1}}=\sqrt{\frac{16}{1}}=4 \).

Teacher's Note:
1. Lighter gas is faster.
2. Take the square root of the density ratio.

Answer: (a) 79.19 mL/s

Explanation:
1. \( \frac{r_{N_2}}{r_{gas}}=\sqrt{\frac{2M_{N_2}}{M_{N_2}}}=\sqrt2 \).
2. So \( r_{N_2}=56\sqrt2=79.19 \) mL/s.

Teacher's Note:
1. Nitrogen is lighter, so its rate is more than 56.
2. Multiply by the square root of 2.

Answer: (c) H₂

Explanation:
1. Leakage through pores is effusion, whose rate varies as \( 1/\sqrt M \).
2. H₂ is lightest, so it escapes fastest.

Teacher's Note:
1. Lightest gas escapes first.
2. The source wording said filled first; rewritten to match the intended idea of fastest loss.