The vapour pressure of water at $23^{0} C$ is 19.8 mm. 0.1 mole of glucose is dissolved in 178.2 g of water. What is the vapour pressure (in mm) of the resultant solution?

19.0

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19.602

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19.402

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19.202

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Solution

The correct option is D $19.602$
$A \to$ solvent
$B \to$ solute
$P_{A}^{O} = 19.8$
$n_{A} = \frac{178.2}{18}$ $n_{B} = 0.1$
$n_{A} = 9.9$
$X_{A} = \frac{n_{A}}{n_{A} + n_{B}}$
$X_{A} = \frac{9.9}{9.9 + 0.1} = \frac{9.9}{10} = 0.99$
$\frac{P_{A}}{P_{A}^{O}} = X_{A}$
$P_{A} = 19.8 \times 0.99$

$P_{A} = 19.602$

Hence, the correct option is $B$

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