JEE Chemistry States Of Matter MCQs Set 12

Answer: (a) the same

Explanation:
1. Average translational KE is \(\frac32 kT\), independent of mass.
2. So both have the same value.

Teacher's Note:
1. Mass affects the speed but not the average KE.

Answer: (a) 120 mL

Explanation:
1. Each acetone molecule has 3 C atoms.
2. 40 mL gives \(3 \times 40 = 120\) mL CO\(_2\).

Teacher's Note:
1. One volume of acetone gives three volumes of CO\(_2\).

Answer: (d) C\(_5\)H\(_{12}\)

Explanation:
1. CO\(_2\)/HC = 5 so x = 5.
2. H\(_2\)O/HC = 6, so y = 12. The formula is C\(_5\)H\(_{12}\).

Teacher's Note:
1. Volume ratios equal mole ratios in the same conditions.

Answer: (d) C\(_2\)H\(_4\)

Explanation:
1. CO\(_2\) = 15 mL, so x = 15/7.5 = 2.
2. O\(_2\) left = 13.5 mL, used = 22.5 mL, i.e. 3 per mL: \(x + y/4 = 3\), so y = 4.
3. The formula is C\(_2\)H\(_4\).

Teacher's Note:
1. KOH absorbs CO\(_2\) and pyrogallol absorbs O\(_2\).

Answer: (b) C\(_2\)H\(_6\)

Explanation:
1. For C\(_n\)H\(_{2n+2}\): O\(_2\) = (3n+1)/2, CO\(_2\) = n.
2. \((3n+1)/(2n) = 7/4\) gives \(12n + 4 = 14n\), so n = 2.

Teacher's Note:
1. Alkanes follow the general formula C\(_n\)H\(_{2n+2}\).

Answer: (d) 2510 L

Explanation:
1. Moles of C\(_4\)H\(_{10}\) = 1000/58 = 17.24.
2. O\(_2\) needed = 6.5 per mole = 112.1 mol.
3. Volume = 112.1 \(\times\) 22.4 = 2510 L.

Teacher's Note:
1. \(\ce{C4H10 + 6.5O2 -> 4CO2 + 5H2O}\).

Answer: (d) 30 mL

Explanation:
1. O\(_3\) = 60 mL.
2. \(\ce{2O3 -> 3O2}\): 60 mL gives 90 mL.
3. Increase = 30 mL.

Teacher's Note:
1. Increase = half of the ozone volume.

Answer: (d) 4 : 1

Explanation:
1. Let CH\(_4\) = x, CO = y. O\(_2\) = 2x + 0.5y = 1.7(x + y).
2. \(0.3x = 1.2y\), so x : y = 4 : 1.

Teacher's Note:
1. \(\ce{CH4 + 2O2 -> CO2 + 2H2O}\) and \(\ce{2CO + O2 -> 2CO2}\).

Answer: (d) 0.75

Explanation:
1. Reacting volume = 30 mL in the ratio 2 : 1, i.e. 20 mL H\(_2\) and 10 mL O\(_2\).
2. H\(_2\) originally = 20 + 10 = 30 mL.
3. Mole fraction = 30/40 = 0.75.

Teacher's Note:
1. Add the leftover hydrogen back.

Answer: (a) 126.67 m

Explanation:
1. Maximum volume = 600 mL, so burst pressure = \(760 \times 500/600 = 633.33\) mm.
2. Fall in pressure = 126.67 mm, i.e. 126.67 \(\times\) 100 cm = 126.67 m.

Teacher's Note:
1. Boyle's law at constant temperature.

Answer: (a) 1.92 atm

Explanation:
1. Initial moles = 0.05. 30% dissociated: 0.015 mol gives 0.030 mol atoms.
2. Total = 0.035 + 0.030 = 0.065 mol.
3. \(P = 0.065 \times 0.0821 \times 1800/5 = 1.92\) atm.

Teacher's Note:
1. Each dissociated molecule gives two atoms.

Answer: (c) 0.5714 atm

Explanation:
1. Equal pressure: \(n_{27} \times 300 = n_{127} \times 400\) with total 0.70.
2. \(n_{27} = 0.40\), \(n_{127} = 0.30\).
3. \(V = 0.35 \times 0.0821 \times 300/0.5 = 17.24\) L, so \(P = 0.40 \times 0.0821 \times 300/17.24 = 0.5714\) atm.

Teacher's Note:
1. Moles move to the cooler flask.

Answer: (a) 0.4 mol and 0.3 mol

Explanation:
1. Equal pressure and volume give \(n \propto 1/T\).
2. \(n_1 : n_2 = 400 : 300 = 4 : 3\).
3. \(n_1 = 0.4\), \(n_2 = 0.3\).

Teacher's Note:
1. The hotter flask holds fewer moles.

Answer: (b) 4.3

Explanation:
1. For O\(_2\), 1 L takes 600 s.
2. \(M_{mix} = 32 \times (311/600)^2 = 8.6\).
3. Vapour density = 8.6/2 = 4.3.

Teacher's Note:
1. Compare equal volumes.

Answer: (c) 46.6

Explanation:
1. \(M_{mix} = 32 \times (234/224)^2 = 34.92\).
2. \(0.8 \times 32 + 0.2M = 34.92\) gives \(M = 46.6\).

Teacher's Note:
1. Average molar mass is weighted by mole fraction.

Answer: (b) 81.1 cm

Explanation:
1. Distances are proportional to rates: \(d_{HCl}/d_{NH_3} = \sqrt{17/36.5} = 0.6825\).
2. \(d_{HCl} = 200 \times 0.6825/1.6825 = 81.1\) cm.

Teacher's Note:
1. The lighter NH\(_3\) travels farther.

Answer: (d) 71 g/mol

Explanation:
1. \(M = 40 \times (60/45)^2 = 40 \times 16/9\).
2. \(M = 71.1\) g/mol.

Teacher's Note:
1. Time is proportional to \(\sqrt{M}\).

Answer: (b) C\(_2\)H\(_2\)

Explanation:
1. CO\(_2\) = 20 mL gives x = 2.
2. Condensed steam = 10 mL gives y/2 = 1, so y = 2.
3. The formula is C\(_2\)H\(_2\).

Teacher's Note:
1. Contraction on cooling equals the volume of water vapour.

Answer: (b) 346 mL

Explanation:
1. Propane = 73 mL gives \(3 \times 73 = 219\) mL CO\(_2\).
2. The remaining 127 mL (CH\(_4\) and CO) each give 1 CO\(_2\) per molecule: 127 mL.
3. Total = 346 mL.

Teacher's Note:
1. Count the carbon atoms per molecule.

Answer: (a) \(99.5^\circ\)C

Explanation:
1. At constant V, \(P/T\) is constant.
2. \(T = 300 \times 14.9/12 = 372.5\) K = \(99.5^\circ\)C.

Teacher's Note:
1. 372.5 K is the same temperature as 99.5\(^\circ\)C.

Answer: (b) I and III only

Explanation:
1. Rate ratio = \(\sqrt{M/4}\).
2. CO: \(\sqrt{7} = 2.65\) (true); CO\(_2\): \(\sqrt{11} = 3.3\) (false); SO\(_2\): \(\sqrt{16} = 4\) (true).

Teacher's Note:
1. Convert multi-statement items by checking each separately.

Answer: (b) II and III only

Explanation:
1. Same T gives equal KE, so lighter He has the higher speed and hits the wall more often.
2. Partial pressures are equal because moles are equal.
3. So II and III are correct.

Teacher's Note:
1. Equal moles in the same volume give equal partial pressures.

Answer: (c) I, III and IV

Explanation:
1. Same T gives the same KE per molecule; equal V, T, P gives equal moles, so equal total KE.
2. Speeds differ because M differs.
3. N\(_2\) (28) is lighter than CO\(_2\) (44), so its density is lower.

Teacher's Note:
1. Average KE depends only on T.

Answer: (c) Total pressure becomes 25 atm and \(p_C\) = 5 atm

Explanation:
1. Pressure is proportional to moles at fixed V and T.
2. 4 mol gave 20 atm, so 5 mol gives 25 atm and 1 mol of C gives 5 atm.
3. Partial pressures of A and B do not change.

Teacher's Note:
1. Each gas in a mixture behaves independently.