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Q.

At 301 K the vapour pressure of pure water is 3.780 kPa. When a certain weight of solute of molar mass 342 is dissolved in 1000 gms of water, the vapour pressure is lowered ta 3.776 kPa. The molality of the solute is

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a

0.263

b

0.315

c

0.206

d

0.354

answer is C.

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Detailed Solution

$\begin{array}{l} Mole fraction of solute = \frac{3 .78 - 3.766}{3 .78} = 0 .0037 \\ Moles of water = 1 - 0 .0037 = 0 .9963; \\ Weight of water = 0 .9963 \times 18 gm \\ Molality of solute = \frac{0 .0037 \times 1000}{0 .9963 \times 18} = 0 .206 . \end{array}$

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