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Liquids in Liquids

The vapour pr...

Question

The vapour pressure of pure water is 23.8 mm Hg at $25^{0} C$ What is the vapour pressure of 2.50 molal $C_{6} H_{12} O_{6}$ .

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Solution

First convert the molality to a mole fraction
$\Rightarrow 2.50 m$ $C_{6} H_{12} O_{6} = 2.50$ mol/ $1 k g$ of $H_{2} O$
$\Rightarrow 1000 g / 18.015 = 55.51$ mol
$\Rightarrow 55.51 + 2.50 = 58.01$ mol
Mole fraction of $H_{2} O = \frac{55.81}{58.01} = 0.9569$
Raoult's Law:
$P_{s o l u t i o n} = (Y s o l v e n t) (P^{o} s o l v e n t)$
$= (0.9569) (23.8) = 22.8 m m H g$ .

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