JEE Chemistry States Of Matter MCQs Set 20

Answer: (c) 14.14 L

Explanation:
1. Rate of diffusion = volume/time. For SO2: \( r_1=20/120 \) L s\(^{-1}\).
2. Graham's law: \( \frac{r_{O_2}}{r_{SO_2}}=\sqrt{\frac{64}{32}}=\sqrt{2} \).
3. \( r_{O_2}=\frac{20}{120}\sqrt{2}=0.2357 \) L s\(^{-1}\), so in 60 s the volume = \( 0.2357\times60=14.14 \) L.

Teacher's Note:
1. Rate is inversely proportional to the square root of molar mass at the same T and P.
2. The lighter gas diffuses more in the same time.

Answer: (a) \( \ce{N2O} \)

Explanation:
1. \( \frac{r_X}{r_{PH_3}}=\sqrt{\frac{34}{M_X}}=0.88 \).
2. \( M_X=\frac{34}{0.88^2}=43.9\approx44 \) g mol\(^{-1}\).
3. N2O has molar mass 44 (CO2 would too, but it is not listed).

Teacher's Note:
1. Square both sides of Graham's law to get the molar mass.
2. C2H6 = 30, CO = 28, NO2 = 46, N2O = 44.

Answer: (d) At 6.1 K

Explanation:
1. \( P=\frac{n}{V}RT=CRT \), so \( T=\frac{P}{CR}=\frac{1}{2\times0.0821}=6.1 \) K.
2. At STP the concentration is 1/22.4 mol per litre, not 2.

Teacher's Note:
1. For an ideal gas, \( P=CRT \).
2. The answer is a very low temperature, where real gases would not behave ideally, but the question is for an ideal gas.

Answer: (b) 1

Explanation:
1. \( Z=\frac{PV}{nRT} \).
2. For an ideal gas \( PV=nRT \), so Z = 1 at all temperatures and pressures.

Teacher's Note:
1. Z < 1 means attractive forces dominate; Z > 1 means repulsive forces dominate.

Answer: (b) 1 : 25

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

Teacher's Note:
1. Faster diffusion means lower density (and lower molar mass).
2. Square the rate ratio.

Answer: (d) 1.253 L\(^2\) mol\(^{-2}\) atm

Explanation:
1. \( Z=\frac{PV}{RT}=0.5 \), so \( V=\frac{0.5\times0.0821\times273}{100}=0.1121 \) L.
2. With b negligible: \( \left(P+\frac{a}{V^2}\right)V=RT \), so \( a=V(RT-PV) \).
3. \( a=0.1121\,(22.41-11.21)=1.256\approx1.25 \) L\(^2\) mol\(^{-2}\) atm.

Teacher's Note:
1. If b is neglected the equation reduces to \( P+\frac{a}{V^2}=\frac{RT}{V} \).
2. Z < 1 shows attraction (the a term) dominates.

Answer: (a) 0.31

Explanation:
1. Rate of effusion is proportional to \( \frac{x}{\sqrt{M}} \) (partial pressure over \( \sqrt{M} \)).
2. \( \frac{x_{CH_4}}{\sqrt{16}}=\frac{x_{HBr}}{\sqrt{81}} \), so \( \frac{x_{CH_4}}{x_{HBr}}=\frac{4}{9} \).
3. \( x_{CH_4}=\frac{4}{13}=0.31 \).

Teacher's Note:
1. For a mixture, the effusion rate depends on both composition and molar mass.
2. The fraction of the lighter gas is smaller when the rates are equal.

Answer: (c) 20 seconds, oxygen

Explanation:
1. Time is proportional to \( \sqrt{M} \): \( t=5\sqrt{M/2} \).
2. O2: \( 5\sqrt{16}=20 \) s. He: \( 5\sqrt{2}=7.1 \) s. CO: \( 5\sqrt{14}=18.7 \) s. CO2: \( 5\sqrt{22}=23.5 \) s.
3. Only the oxygen pairing (20 s) is correct.

Teacher's Note:
1. For the same volume, time is directly proportional to the square root of molar mass.

Answer: (c) A gas obeying \( \left(P+\frac{a}{V^2}\right)(V-b)=RT \)

Explanation:
1. Liquefaction needs intermolecular attraction, which the term \( a/V^2 \) represents.
2. The ideal gas law has no attraction and \( P(V-b)=RT \) has only repulsion (size), so neither can liquefy.
3. Obeying the Maxwell distribution says nothing about attraction.

Teacher's Note:
1. The a term in the van der Waals equation accounts for attractive forces.

Answer: (a) \( \left[\frac{3}{4\pi}\times10^{-23}\right]^{1/3} \) cm

Explanation:
1. \( V_c=3b \), so \( b=0.024 \) L mol\(^{-1}\) = 24 cm\(^3\) mol\(^{-1}\).
2. \( b=4N_A\cdot\frac{4}{3}\pi r^3 \), so \( r^3=\frac{3b}{16\pi N_A}=\frac{3\times24}{16\pi\times6.022\times10^{23}} \).
3. \( r^3=\frac{3}{4\pi}\times\frac{6}{6.022\times10^{23}}\approx\frac{3}{4\pi}\times10^{-23} \) cm\(^3\), so \( r=\left[\frac{3}{4\pi}\times10^{-23}\right]^{1/3} \) cm.

Teacher's Note:
1. b is four times the actual volume of the molecules in one mole.
2. Use \( V_c=3b \) first.

Answer: (d) 2.71 L\(^2\) atm mol\(^{-2}\)

Explanation:
1. \( V_c=\frac{44}{0.45}=97.8 \) cm\(^3\) mol\(^{-1}\) = 0.0978 L mol\(^{-1}\), so \( b=V_c/3=0.0326 \) L.
2. \( T_c=\frac{8a}{27Rb}=300 \) K gives \( a=\frac{27RbT_c}{8} \).
3. \( a=\frac{27\times0.0821\times0.0326\times300}{8}=2.71 \) L\(^2\) atm mol\(^{-2}\).

Teacher's Note:
1. Use \( V_c=3b \) and \( T_c=\frac{8a}{27Rb} \).
2. Critical density = molar mass / critical volume.

Answer: (b) \( PV-Pb=RT \)

Explanation:
1. H2 and He have very weak attractions, so the constant a is very small and the term \( a/V^2 \) can be dropped.
2. The equation \( \left(P+\frac{a}{V^2}\right)(V-b)=RT \) becomes \( P(V-b)=RT \), i.e. \( PV-Pb=RT \).

Teacher's Note:
1. Because the a term is negligible for H2 and He, Z = 1 + Pb/RT, which is greater than 1.