1 mole of a non-volatile solid is dissolved in 200 moles of water. The solution in taken to a temperature T K (lower than freezing point of solution) to cause ice formation. After removal of ice the

Question

1 mole of a non-volatile solid is dissolved in 200 moles of water. The solution in taken to a temperature $T_{K}$ (lower than freezing point of solution) to cause ice formation. After removal of ice the remaining solution is taken to 373 K where vapour pressure is observed to be 740 mm of Hg . Identify the correct options:

Given data: $K_{f}$ (H₂O) = 2 K kg mol⁻¹ and normal boiling point of H₂O = 373 K

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Here's a refined solution:

Initial solution: 1 mole solute, 200 moles H₂O.

Freezing point depression:

$Δ T_{f} = K_{f} \times m = 2 \times \frac{1 m o l}{200 \times 18 \times 1 0 ^{- 3} k g} = 2 \times \frac{1000}{3600} = \frac{10}{18} = 0.55 5^{\circ} C$

So, $T_{f} = - 0.55 5^{\circ} C$ . (Option C is correct)

After ice removal, at 373 K (boiling point of water), vapor pressure $P = 740 mm H g$ .

Pure water vapor pressure $P_{0} = 760 mm H g$ .

Raoult's Law: $\frac{P _{0} - P}{P _{0}} = X_{so l u t e}$

$\frac{760 - 740}{760} = \frac{20}{760} = \frac{1}{38}$

$X_{so l u t e} = \frac{n _{so l u t e}}{n _{so l u t e} + n _{H_{2} O, f ina l}} = \frac{1}{1 + n _{H_{2} O, f ina l}} = \frac{1}{38}$

$1 + n_{H_{2} O, f ina l} = 38 ⟹ n_{H_{2} O, f ina l} = 37 m o l es$ .

Initial moles of water = 200 moles.

Moles of ice formed = $200 - 37 = 163 m o l es$ . (Option A is correct)

Relative lowering of vapor pressure of final solution = $\frac{1}{38}$ . (Option D is incorrect)

The temperature to which the original solution was cooled is $T_{K}$ . This is the freezing point of the solution, which is $273.15 - 0.555 = 272.595 K$ . (Option B is incorrect)

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Doubts & Discussion

JEE Advanced 2027 (Class 12)