JEE Chemistry Solutions MCQs Set 17

Answer: (b) ΔS mixing = 0

Explanation:
1. For an ideal solution ΔH(mix) = 0, ΔV(mix) = 0 and Raoult's law is obeyed.
2. The entropy of mixing is positive (the solution is more disordered), so ΔS(mix) is not zero.

Teacher's Note:
1. The mixing of two ideal liquids is driven by an increase in entropy.

Answer: (d) 2 M sucrose solution

Explanation:
1. 5.8% means 58 g per litre, \( 58/58.5 \approx 1 \) M of NaCl.
2. With i = 2 the effective concentration is about 2 M, close to a 2 M sucrose solution.

Teacher's Note:
1. The effective particle concentration of NaCl is double its molarity.

Answer: (a) 3

Explanation:
1. \( Ba(OH)_2\to Ba^{2+}+2OH^- \) gives 3 ions, so i = 3.

Teacher's Note:
1. A strong electrolyte in dilute solution is almost completely dissociated.

Answer: (b) -0.654 °C

Explanation:
1. \( m = \dfrac{0.18}{0.512} = 0.3516 \).
2. \( \Delta T_f = 1.86\times0.3516 = 0.654 \), so the freezing point is \( -0.654 \) °C.

Teacher's Note:
1. The same solution gives both the boiling point elevation and the freezing point depression.

Answer: (b) C2H5Cl and C6H5OH

Explanation:
1. Ideal solutions form between molecules of similar structure and polarity (benzene-toluene, chlorobenzene-bromobenzene, bromoethane-iodoethane).
2. Phenol is strongly hydrogen-bonded, whereas chloroethane is not, so this pair deviates from ideality.

Teacher's Note:
1. Hydrogen bonding in one component usually leads to non-ideal behaviour.

Answer: (b) (NH4)3PO4

Explanation:
1. The osmotic pressure is proportional to the number of ions: AgNO3 gives 2, Na2SO4 gives 3, (NH4)3PO4 gives 4 and MgCl2 gives 3.
2. (NH4)3PO4 gives the most ions.

Teacher's Note:
1. For equal molarity, compare the number of ions per formula unit.

Answer: (d) ΔG(mix) = 0

Explanation:
1. For an ideal solution, the deviation from Raoult's law, ΔH(mix), ΔV(mix) and ΔU(mix) are all zero.
2. ΔG(mix) = −TΔS(mix) is negative, so ΔG(mix) = 0 is incorrect.

Teacher's Note:
1. Since ΔH = 0 and ΔS > 0, the free energy of mixing is negative.

Answer: (a) 800.00 g

Explanation:
1. \( m = \dfrac{6}{1.86} = 3.226 \) mol/kg, so moles in 4 kg \( = 12.9 \).
2. Mass \( = 12.9\times62 = 800 \) g.

Teacher's Note:
1. Find the molality needed for the desired depression, then scale up.

Answer: (d) shows a positive deviation from Raoult's law

Explanation:
1. Positive deviation means a higher vapour pressure than predicted, so the boiling point is lower.
2. Such a mixture is a minimum-boiling azeotrope.

Teacher's Note:
1. Positive deviation gives a minimum-boiling azeotrope. Negative deviation gives a maximum-boiling azeotrope.

Answer: (c) 1 M alum solution

Explanation:
1. Glucose and urea give i = 1 and NaCl gives i = 2.
2. Alum, \( K_2SO_4\cdot Al_2(SO_4)_3\cdot24H_2O \), gives 2 K+, 2 Al3+ and 4 SO4(2-) = 8 ions per formula unit, the greatest.

Teacher's Note:
1. Count the total number of ions released per formula unit.

Answer: (d) -3.6 °C, 101 °C

Explanation:
1. \( m = \dfrac{0.5}{0.5} = 1 \) molal, \( i = 2 \).
2. \( \Delta T_f = 2\times1.8\times1 = 3.6 \), so \( T_f = -3.6 \) °C.
3. \( \Delta T_b = 2\times0.5\times1 = 1 \), so \( T_b = 101 \) °C.

Teacher's Note:
1. The freezing point decreases and the boiling point increases. NaCl has i = 2.

Answer: (d) 93 g

Explanation:
1. \( m = \dfrac{2.8}{1.86} = 1.505 \) mol/kg.
2. Mass \( = 1.505\times62 = 93.3 \) g.

Teacher's Note:
1. Molar mass of C2H6O2 is 62 g/mol.

Answer: (c) Refractive index

Explanation:
1. Colligative properties depend only on the number of solute particles.
2. Refractive index depends on the nature of the substance, so it is not colligative.

Teacher's Note:
1. The four colligative properties are the three listed and the elevation in boiling point.

Answer: (b) The osmotic pressure of Na2SO4 is more than that of the NaCl solution

Explanation:
1. NaCl gives 2 ions (i = 2) and Na2SO4 gives 3 ions (i = 3).
2. At equal molarity, Na2SO4 has the larger osmotic pressure.

Teacher's Note:
1. The more ions, the higher the osmotic pressure.

Answer: (b) 256

Explanation:
1. \( M = \dfrac{K_f\times w\times1000}{\Delta T_f\times W} = \dfrac{5.12\times1\times1000}{0.4\times50} = 256 \) g/mol.

Teacher's Note:
1. Use the standard molar mass formula with masses in grams.

Answer: (c) \( 5.16\times10^6 \)

Explanation:
1. \( \pi = \dfrac{C_{(g/L)}}{M}RT \), and \( C_{(g/L)} = 1000\,C_{(g/cm^3)} \).
2. Slope \( = \dfrac{1000RT}{M} \Rightarrow M = \dfrac{1000\times0.0821\times293}{4.65\times10^{-3}} \approx 5.17\times10^6 \).

Teacher's Note:
1. The slope of a π versus C plot gives RT/M, a standard way to find the molar mass of a polymer.

Answer: (c) the strength (concentration) of the solution

Explanation:
1. \( P^0-P_s = P^0x_{solute} \).
2. So the lowering is proportional to the mole fraction (strength) of the solute, not to its nature.

Teacher's Note:
1. Colligative properties depend on how many particles are present, not on what they are.

Answer: (c) 0.1 M Al2(SO4)3

Explanation:
1. Al2(SO4)3 gives 5 ions: effective 0.5 M. BaCl2 and Na2SO4 each give 3 ions: effective 0.3 M.
2. The mixture of (2) and (3) also has an effective concentration of 0.3 M.

Teacher's Note:
1. Compare the total ion concentration i × M for each solution.

Answer: (a) 0.04 L

Explanation:
1. Moles needed \( = 0.4\times2.5 = 1 \) mol, which is 32 g of methanol.
2. Volume \( = \dfrac{32\ \text{g}}{800\ \text{g/L}} = 0.04 \) L.

Teacher's Note:
1. Density 0.8 kg/L is 800 g per litre.

Answer: (c) 0.16 mol L⁻¹

Explanation:
1. \( \pi = iCRT \Rightarrow C = \dfrac{7.8}{2\times0.0831\times310} = 0.151 \) mol/L.
2. The nearest option is 0.16 mol/L.

Teacher's Note:
1. NaCl gives two particles, so the molarity needed is half of what a non-electrolyte would require.

Answer: (a) Raoult

Explanation:
1. This is Raoult's law for solutions of non-volatile solutes.

Teacher's Note:
1. Henry's law is for gases in liquids, Dalton's law is for gas mixtures.

Answer: (c) AgNO3

Explanation:
1. Number of ions: BaCl2 3, AgNO3 2, Na2SO4 3, (NH4)3PO4 4.
2. AgNO3 gives the fewest ions, so it has the minimum osmotic pressure.

Teacher's Note:
1. The fewest particles give the lowest colligative effect.

Answer: (c) \( \dfrac{\text{moles of solute}}{\text{moles of solvent}} \)

Explanation:
1. \( \dfrac{P^0-P_s}{P^0} = \dfrac{n}{n+N} \approx \dfrac{n}{N} \) for a dilute solution.
2. So it depends on the ratio of moles of solute to moles of solvent (increasing with it).

Teacher's Note:
1. Of the given forms, only the first has the solute in the numerator.

Answer: (c) it has a very high cryoscopic constant

Explanation:
1. Camphor has a very large molal depression constant (Kf about 40 K kg/mol).
2. A large Kf gives a large, easily measured depression even for small amounts of solute.

Teacher's Note:
1. A high cryoscopic constant means a large depression for each mole of solute.

Answer: (d) \( \dfrac{P^0-P_s}{P_s} = \dfrac{n}{n+N} \)

Explanation:
1. Raoult's law: \( \dfrac{P^0-P_s}{P^0} = \dfrac{n}{n+N} \). Hence \( \dfrac{P^0-P_s}{P_s} = \dfrac{n}{N} \).
2. Option (3) puts \( n+N \) in the denominator, so it is incorrect. The other three follow correctly.

Teacher's Note:
1. Check each form by taking the standard result and rearranging it.