The vapour pressure of pure water is 760 mm at $25^{0} C$ .The vapour pressure of solution containing 1(m) solution of glucose will be:

761.0 mm

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746.5 mm

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734.5 atm

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746.5 Pa

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Solution

The correct option is D 746.5 mm
Vapor pressure
$\frac{P^{o} - P}{P^{o}} = \frac{W_{2}}{M_{2}} \times \frac{M_{1}}{W_{1}}$

$P^{o} = 760 m m$

$M_{1} = 18$

Molality $(m) = 1 m = \frac{m o l e o f s o l u t e}{(w e i g h t) k g o f s o l v e n t} = \frac{W_{2} / M_{2}}{W_{i} m g m}$

$\frac{760 - P}{760} = 1 \times 10^{- 3} \times 18$

$P = 746.5 m m$

Hence, the correct option is $B$

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