The mass of sucrose to be added to 300 g of water to lower its vapour pressure by 1.0 mmHg at 25°C is (Given: p∗( water )=23.8mmHg

The mass of sucrose to be added to 300 g of water to lower its vapour pressure by 1.0 mmHg at 25°C is (Given: $p^{*} (water) = 23.8 mmHg)$

A
249.9 g
B
329.4 g
C
215.2 g
D
342.2 g

Solution:

Since. $p = x_{1} p_{1}$ we get

$x_{1} = \frac{p}{p_{1}^{*}} = \frac{23.8 - 1.0}{23.8} = 0.958$ Now $x_{1} = \frac{n_{1}}{n_{1} + n_{2}} = \frac{(300 / 18) mol}{(300 / 18) mol + (m / 342 g {mol}^{- 1})}$

Equating these two, we get

$\frac{(300 / 18)}{(300 / 18) + (m / 342 g)} = 0.958$ or $m = \frac{{(300 / 18) (1 - 0.958)} (342 g)}{0.958} = 249.89 g$

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