JEE Chemistry States Of Matter MCQs Set 11

Answer: (b) 4373 kg

Explanation:
1. Pay load = mass of air displaced - mass of balloon - mass of helium.
2. \(5108 - 50 - 685 = 4373\) kg.

Teacher's Note:
1. Buoyancy equals the weight of the displaced air.

Answer: (d) 2 : 1

Explanation:
1. \(\ce{N2 + 3H2 -> 2NH3}\): 4 mol become 2 mol.
2. The ratio of volumes = 4 : 2 = 2 : 1.

Teacher's Note:
1. At constant T and P, volume is proportional to moles.

Answer: (a) 12 K

Explanation:
1. \(P = (n/V)RT\), so \(T = P/(CR) = 1/0.082 = 12.2\) K.

Teacher's Note:
1. STP gives 0.0446 mol/L, not 1 mol/L.
2. The state is physically unrealistic for a real gas but valid for an ideal gas.

Answer: (a) B\(_2\)H\(_6\)

Explanation:
1. \(n = PV/RT = 0.820/(0.0820 \times 276) = 0.03623\) mol.
2. \(M = 1/0.03623 = 27.6\) g/mol.
3. B\(_2\)H\(_6\) = 21.6 + 6 = 27.6.

Teacher's Note:
1. Find the molar mass from PV = (m/M)RT.
2. Compare with each candidate.

Answer: (d) 4

Explanation:
1. \(P = dRT/M\), so \(P_A/P_B = (d_A/d_B)(M_B/M_A)\).
2. \(= 2 \times 2 = 4\).

Teacher's Note:
1. Flask volume does not matter once the densities are given.

Answer: (b) 75

Explanation:
1. Equal V, T and P means equal moles.
2. Moles of ethane = 15/30 = 0.5, so \(M(X_2) = 75/0.5 = 150\).
3. Vapour density = M/2 = 75.

Teacher's Note:
1. Vapour density is half of the molar mass.

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

Explanation:
1. \(d \propto P/T\).
2. Values: STP 1/273; 2/273; 1/546; 2/546.
3. The largest is 2 atm at 273 K.

Teacher's Note:
1. High pressure and low temperature give the highest density.

Answer: (a) 12.7 mL

Explanation:
1. \(V_2 = V_1(P_1/P_2)(T_2/T_1) = 2 \times 6 \times 298/281\).
2. \(= 12.73\) mL.

Teacher's Note:
1. Convert both temperatures to kelvin.

Answer: (c) 15 : 16

Explanation:
1. For mass m: \(n_{H_2} = m/2\), \(n_{C_2H_6} = m/30\).
2. Fraction of H\(_2\) = \((1/2)/(1/2 + 1/30) = 15/16\).

Teacher's Note:
1. Pressure fraction equals mole fraction.

Answer: (c) 0.8 bar

Explanation:
1. In 100 g: H\(_2\) = 10 mol, O\(_2\) = 80/32 = 2.5 mol.
2. \(x_{H_2} = 10/12.5 = 0.8\).
3. \(p = 0.8\) bar.

Teacher's Note:
1. Weight percent is not mole fraction.
2. Light gases have large mole fractions.

Answer: (c) 2 atm

Explanation:
1. 48 g dimer = 0.5 mol; 48 g monomer = 1 mol; total 1.5 mol.
2. \(P = 1.5 \times 0.0821 \times 546/33.6 = 2.0\) atm.

Teacher's Note:
1. Count molecules of each form, not atoms.
2. 50% by weight is not 50% by moles.

Answer: (c) 0.2

Explanation:
1. Reacted gas = 1.0 - 0.4 = 0.6 atm-equivalent, in the ratio 2 H\(_2\) : 1 O\(_2\).
2. O\(_2\) = 0.2, so H\(_2\) = 0.8.
3. \(x_{O_2} = 0.2\).

Teacher's Note:
1. \(\ce{2H2 + O2 -> 2H2O}\).

Answer: (a) 150 cm\(^3\)

Explanation:
1. Boyle's law: \(P \times 100 = 0.4P \times (100 + V_B)\).
2. \(100 + V_B = 250\), so \(V_B = 150\) cm\(^3\).

Teacher's Note:
1. Total volume after opening is 250 cm\(^3\).

Answer: (d) drops by 57%

Explanation:
1. Let \(P_B = 1\), \(P_A = 20\). Total \(PV = 2000 + 150 = 2150\).
2. Final \(P = 2150/250 = 8.6\).
3. Drop = \((20 - 8.6)/20 = 57\%\).

Teacher's Note:
1. Total PV is conserved at constant T.

Answer: (a) 1.125 atm

Explanation:
1. Moles are proportional to PV: O\(_2\) = 100, NO = 450 units.
2. \(\ce{2NO + O2 -> 2NO2}\): O\(_2\) is limiting, using 200 NO; NO left 250, NO\(_2\) 200.
3. Total 450 units in 400 mL gives \(P = 450/400 = 1.125\) atm.

Teacher's Note:
1. Find the limiting reagent first.
2. The total moles fall by 100 units.

Answer: (a) CO\(_2\) > SO\(_2\) > SO\(_3\) > PCl\(_3\)

Explanation:
1. Rate is inversely proportional to \(\sqrt{M}\).
2. Molar masses: CO\(_2\) 44, SO\(_2\) 64, SO\(_3\) 80, PCl\(_3\) 137.5.
3. Lightest diffuses fastest.

Teacher's Note:
1. Memorise approximate molar masses.

Answer: (a) 14.14 dm\(^3\)

Explanation:
1. Rate of SO\(_2\) = 20/60 = 1/3 dm\(^3\)/s.
2. \(r_{O_2}/r_{SO_2} = \sqrt{64/32} = \sqrt{2}\), so \(r_{O_2} = 0.4714\).
3. Volume in 30 s = 14.14 dm\(^3\).

Teacher's Note:
1. Rate = volume/time.

Answer: (c) 20 s

Explanation:
1. \(t \propto \sqrt{M}\).
2. \(t_{O_2} = 5 \times \sqrt{32/2} = 5 \times 4 = 20\) s.

Teacher's Note:
1. The molar mass ratio 16 gives a time ratio 4.

Answer: (c) \(1 : \sqrt{14} : \sqrt{7}\)

Explanation:
1. Rate \(\propto 1/\sqrt{M}\): \(1/\sqrt{28} : 1/\sqrt{2} : 1/\sqrt{4}\).
2. Multiply by \(\sqrt{28}\): \(1 : \sqrt{14} : \sqrt{7}\).

Teacher's Note:
1. Write the ratio and clear the roots.

Answer: (c) 480 K

Explanation:
1. \(T_{O_2}/32 = 300/20\).
2. \(T = 480\) K.

Teacher's Note:
1. The heavier gas needs a higher temperature.

Answer: (a) 300 m s\(^{-1}\)

Explanation:
1. \(u_{rms} = \sqrt{3P/d}\).
2. \(= \sqrt{3 \times 1.2 \times 10^5/4} = \sqrt{9 \times 10^4} = 300\) m s\(^{-1}\).

Teacher's Note:
1. From \(PV = \frac{1}{3}mNu^2\) we get \(P = \frac13 d u^2\).

Answer: (d) 8 : 1

Explanation:
1. \(P = \frac{1}{3}(mN/V)u^2\).
2. Ratio = \(2 \times 2^2 = 8\).

Teacher's Note:
1. Pressure depends on the mass times the square of the speed.

Answer: (a) \(6.21 \times 10^{-21}\) J

Explanation:
1. \(KE = \frac{3}{2}kT = 1.5 \times 1.38 \times 10^{-23} \times 300\).
2. \(= 6.21 \times 10^{-21}\) J per molecule.

Teacher's Note:
1. Average KE per molecule does not depend on the gas or its mass.

Answer: (b) 1

Explanation:
1. \(\sqrt{50/2} : \sqrt{800/32} = 5 : 5\).
2. The ratio is 1.

Teacher's Note:
1. Compute T/M for each gas.