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:

1. A

249.9 g

2. B

329.4 g

3. C

215.2 g

4. D

342.2 g

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### Solution:

Since. $p={x}_{1}{p}_{1}$ we get

${x}_{1}=\frac{p}{{p}_{1}^{\ast }}=\frac{23.8-1.0}{23.8}=0.958$ Now ${x}_{1}=\frac{{n}_{1}}{{n}_{1}+{n}_{2}}=\frac{\left(300/18\right)\mathrm{mol}}{\left(300/18\right)\mathrm{mol}+\left(\mathrm{m}/342\mathrm{g}{\mathrm{mol}}^{-1}\right)}$

Equating these two, we get

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

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