1 mole of a non-volatile non-dissociative solid solute is dissolved in 200 moles of water. The solution is taken to a temperature TK (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: $K_{f} (H_{2} O) = 2 K k g m o l^{- 1}$ and normal boiling point of $H_{2} O = 373 K$ .

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Temperature to which original solution was cooled should be $\frac{2000}{37 \times 18} K$

163 moles of ice will be formed.

Relative lowering of vapour pressure of final solution will be $\frac{1}{201}$

Freezing point of original solution should be $- \frac{10}{18}$ ⁰C

answer is A, C.

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Detailed Solution

$\frac{760 - 740}{740} = \frac{1}{^{n} H_{2} O}$
$\Rightarrow n_{H_{2} O} = 37$
$\Rightarrow n_{i c e} = 200 - 37 = 163$
$\Rightarrow Δ T_{f} = 2 \times \frac{1}{37 \times 18} \times 1000 K$
$T_{f} = - (\frac{2000}{37 \times 18})$ ⁰C
c) $T_{f}$ (Original solution) $= 2 \times \frac{1}{200 \times 18} \times 1000$ $= \frac{- 10}{18}$ ⁰C
d) R.L of final solution = $\frac{760 - 740}{760} = \frac{1}{38}$

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