SolutionHard

Question

An amount of 1 mole of a non-volatile solid is dissolved in 200 moles of water. The solution is cooled to a temperature ‘T’ K (lower than the freezing point of the solution) to cause ice formation. After the removal of ice, the remaining solution is heated to 373 K, where the vapour pressure of solution is observed to be 740 mm Hg. Identify the correct information(s) from the following (Kf of water = 2.0 K-kg/mol).

Options

A.163 moles of ice is formed at T K.
B.$T\text{ K = }\left( 273 - \frac{2000}{37 \times 18} \right)\text{ K}$
C.Freezing point of the original solution is $- \frac{10}{18}$°C.
D.Relative lowering of vapour pressure of the final solution is $\frac{1}{38}$.

Solution

(a) $\frac{P^{o} - P}{P} = \frac{n_{1}}{n_{2}} \Rightarrow \frac{760 - 740}{740} = \frac{1}{n_{2}} \Rightarrow n_{2} = 37$

∴ Moles of water separated as ice = 200 – 37 = 163

(b) $\Delta T_{f} = K_{f}.m = 2.0 \times \frac{1}{(37 \times 18)/1000} = \frac{2000}{37 \times 18}K$

$\therefore T = \left( 273 - \frac{2000}{37 \times 18} \right)\text{ K}$

(c) For original solution:

$\Delta T_{f} = K_{f}.m = 2 \times \frac{1}{(200 \times 18)/1000} $$$\therefore\text{ F.P. = }0 - \Delta T_{f} = - \frac{10^{o}}{18}C$$

(d) For final solution: $\frac{P^{o} - P}{P^{o}} = X_{1} = \frac{1}{1 + 37} = \frac{1}{38}$

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