JEE Chemistry Solutions MCQs Set 22

Answer: (b) \( \frac{i-1}{N-1} \)

Explanation:
1. Start with 1 mol of electrolyte, of which \( \alpha \) dissociates. Moles after dissociation = \( 1 - \alpha + N\alpha = 1 + (N-1)\alpha \).
2. \( i = 1 + (N-1)\alpha \), so \( \alpha = \frac{i-1}{N-1} \).

Teacher's Note:
1. Use \( \alpha = \frac{i-1}{N-1} \) for dissociation. For association use the relation \( \alpha = \frac{1-i}{1-1/n} \).

Answer: (c) it is the water that solidifies first

Explanation:
1. In a dilute solution, the solid that separates at the freezing point is pure solvent (ice).
2. The solute stays in the remaining liquid, which becomes more concentrated.

Teacher's Note:
1. This is why the freezing point of a solution keeps falling as freezing proceeds.

Answer: (d) neither side

Explanation:
1. A semipermeable membrane allows only solvent molecules to pass, not ions.
2. Neither Fe\(^{3+}\) nor CNS\(^-\) can cross, so they never meet and no red complex forms.

Teacher's Note:
1. Osmosis moves only solvent. Solute particles stay on their own side.

Answer: (a) the number of solute molecules is increased

Explanation:
1. \( \pi = \frac{n}{V}RT \).
2. Osmotic pressure rises with the number of solute particles, rises with temperature and falls when the volume increases.

Teacher's Note:
1. Osmotic pressure is a colligative property, so it depends on the number of solute particles.

Answer: (c) A, C and D only

Explanation:
1. A is true: a non-volatile solute depresses the freezing point.
2. B is false: dilution increases the degree of dissociation of a weak electrolyte.
3. C is true: \( \Delta H_{sol} = \) lattice energy \(-\) hydration energy, which is negative when hydration energy is larger.
4. D is true: \( \Delta_{mix}S > 0 \) for ideal solutions.

Teacher's Note:
1. Dilution favours dissociation (Ostwald's dilution law).

Answer: (d) A, B and D only

Explanation:
1. Solubility of a solid depends on temperature, on the nature of the solute and on the nature of the solvent.
2. Pressure has almost no effect on the solubility of solids in liquids, so C is excluded.

Teacher's Note:
1. Pressure matters for gases dissolving in liquids (Henry's law), not for solids.

Answer: (a) Vapour pressure in container A is more than in container B.

Explanation:
1. NaCl is non-volatile, so its addition lowers the vapour pressure of water.
2. Pure water in A therefore has the higher vapour pressure.

Teacher's Note:
1. A non-volatile solute always lowers the vapour pressure of the solvent.

Answer: (c) Ar < CO\(_2\) < CH\(_4\) < HCHO

Explanation:
1. By Henry's law \( p = K_H x \), so for a given pressure the solubility x is inversely proportional to \( K_H \).
2. The higher the \( K_H \), the lower the solubility.
3. Order of \( K_H \): Ar > CO\(_2\) > CH\(_4\) > HCHO, so solubility increases in the order Ar < CO\(_2\) < CH\(_4\) < HCHO.

Teacher's Note:
1. Large \( K_H \) means a poorly soluble gas.

Answer: (c) The mixture forms a minimum boiling azeotrope.

Explanation:
1. The intermolecular forces in benzene and toluene are of nearly the same strength, so the mixture is ideal.
2. An ideal solution has \( \Delta_{mix}H = 0 \), \( \Delta_{mix}V = 0 \) and obeys Raoult's law.
3. Ideal solutions cannot form azeotropes, so C is not true.

Teacher's Note:
1. Only non-ideal solutions form azeotropes.

Answer: (b) 93 g

Explanation:
1. \( \Delta T_f = 2.8 \) K and ethylene glycol is a non-electrolyte (i = 1).
2. \( m = \frac{2.8}{1.86} = 1.505 \) mol per kg.
3. Molar mass = 62 g/mol, so mass = 1.505 \(\times\) 62 = 93.3 g.

Teacher's Note:
1. Ethylene glycol is the non-electrolyte used as antifreeze.

Answer: (c) 0.086

Explanation:
1. 5.2 mol of CH\(_3\)OH are present in 1000 g of water, which is \( \frac{1000}{18} = 55.56 \) mol.
2. \( x = \frac{5.2}{5.2 + 55.56} = \frac{5.2}{60.76} = 0.086 \).

Teacher's Note:
1. A molal solution gives moles of solute per 1000 g of solvent, so convert the water to moles.

Answer: (b) the same boiling point and the same freezing point

Explanation:
1. \( \Delta T_b = K_b m \) and \( \Delta T_f = K_f m \).
2. For the same solvent \( K_b \) and \( K_f \) are fixed, and for non-electrolytes i = 1.
3. Equal molality therefore gives equal elevation and equal depression.

Teacher's Note:
1. For electrolytes the factor i would also have to be equal.

Answer: (b) Benzene and methanol

Explanation:
1. Adding benzene to methanol breaks the hydrogen bonds between methanol molecules, so the attractions in the mixture are weaker and the vapour pressure is higher than expected.
2. Water with HCl or HNO\(_3\) and acetone with chloroform show negative deviation.

Teacher's Note:
1. Acetone-chloroform forms a weak hydrogen bond, giving negative deviation.

Answer: (c) X is undergoing dissociation in water while Y undergoes no change.

Explanation:
1. \( \Delta T_b = i K_b m \) with the same m and the same solvent.
2. A larger elevation means a larger van't Hoff factor.
3. So X dissociates (i greater than 1) while Y does not.

Teacher's Note:
1. The molar mass does not matter when molality is already given.

Answer: (d) \( \Delta_{mix}S = 0 \)

Explanation:
1. For an ideal solution \( \Delta_{mix}H = 0 \), \( \Delta_{mix}V = 0 \) and Raoult's law holds, \( p_B = x_B p_B^0 \).
2. The entropy of mixing is always positive because the mixture is more disordered, so D is incorrect.

Teacher's Note:
1. Ideal means no enthalpy or volume change, but entropy still increases.

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

Explanation:
1. Acetone molecules break the hydrogen bonds in ethanol and cannot form hydrogen bonds themselves.
2. The attractions in the mixture are weaker, so the vapour pressure is higher than expected, a positive deviation.

Teacher's Note:
1. Acetone with ethanol or methanol gives positive deviation. Acetone with chloroform gives negative deviation.

Answer: (a) 80 torr

Explanation:
1. \( n_A = \frac{20}{40} = 0.5 \) mol and \( n_B = \frac{20}{80} = 0.25 \) mol, so \( x_A = \frac{2}{3} \) and \( x_B = \frac{1}{3} \).
2. \( P = x_A p_A^0 + x_B p_B^0 = \frac{2}{3}(100) + \frac{1}{3}(40) = 66.7 + 13.3 = 80 \) torr.

Teacher's Note:
1. Convert masses to mole fractions before applying Raoult's law.

Answer: (c) 125 g/mol

Explanation:
1. \( \Delta T_b = 80.23 - 80.10 = 0.13 \) K.
2. \( \Delta T_b = K_b \times \frac{w_2 \times 1000}{M_2 \times w_1} \), so \( 0.13 = \frac{2.53 \times 0.640 \times 1000}{M_2 \times 100} \).
3. \( M_2 = \frac{2.53 \times 6.40}{0.13} \approx 125 \) g/mol.

Teacher's Note:
1. Azulene is C\(_{10}\)H\(_8\), molar mass 128, which agrees within the precision of the data.

Answer: (b) 34 kJ/mol

Explanation:
1. \( K_b = \frac{M R T_b^2}{1000 \, \Delta H_{vap}} \), so \( \Delta H_{vap} = \frac{M R T_b^2}{1000 K_b} \).
2. \( T_b = 383.7 \) K, so \( \Delta H_{vap} = \frac{92 \times 8.314 \times (383.7)^2}{1000 \times 3.32} \approx 33.9 \) kJ/mol, which is about 34 kJ/mol.

Teacher's Note:
1. Use the temperature in kelvin and the solvent's molar mass in g/mol with the factor 1000.

Answer: (c) 75%

Explanation:
1. \( i = \frac{\text{normal molar mass}}{\text{observed molar mass}} = \frac{164}{65.6} = 2.5 \).
2. XY\(_2\) gives 3 ions: \( i = 1 + 2\alpha \), so \( \alpha = 0.75 \) or 75%.

Teacher's Note:
1. Observed molar mass is lower than the normal value when the solute dissociates.

Answer: (b) 2

Explanation:
1. Molar mass = 52 + 4(17) + 96 + 35.5 = 251.5 g/mol, so moles = \( \frac{0.010}{251.5} = 3.98 \times 10^{-5} \).
2. \( C = \frac{3.98 \times 10^{-5}}{0.025} = 1.59 \times 10^{-3} \) M.
3. \( i = \frac{\pi}{CRT} = \frac{59.1/760}{1.59 \times 10^{-3} \times 0.0821 \times 298} = 2.0 \), so 2 ions.

Teacher's Note:
1. Convert torr to atm by dividing by 760 before using R = 0.0821 L atm/(mol K).