A solution contains 20 g glucose C6H12O6 and 34.2 g of sucrose C12H22O11 in 108 g water. If the vapour pressure of pure water is 35 mmHg, the decrease in vapour pressure of the solvent will be

# A solution contains 20 g glucose $\left({\mathrm{C}}_{6}{\mathrm{H}}_{12}{\mathrm{O}}_{6}\right)$ and 34.2 g of sucrose $\left({\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{11}\right)$ in 108 g water. If the vapour pressure of pure water is 35 mmHg, the decrease in vapour pressure of the solvent will be

1. A

2. B

3. C

4. D

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### Solution:

Amount of glucose, ${n}_{2}=\frac{{m}_{2}}{{M}_{2}}=\frac{20\mathrm{g}}{180\mathrm{g}{\mathrm{mol}}^{-1}}=\frac{1}{9}\mathrm{mol}$; Amount of sucrose, ${n}_{3}=\frac{{m}_{3}}{{M}_{3}}=\frac{34.2\mathrm{g}}{342\mathrm{g}{\mathrm{mol}}^{-1}}=\frac{1}{10}\mathrm{mol}$

Amount of water, ${n}_{1}=\frac{{m}_{1}}{{M}_{1}}=\frac{108\mathrm{g}}{18\mathrm{g}{\mathrm{mol}}^{-1}}=6\mathrm{mol}$

Mole fraction of solutes in solution,

$x=\frac{{n}_{2}+{n}_{3}}{{n}_{1}+{n}_{2}+{n}_{3}}=\frac{\left[\left(1/9\right)+\left(1/10\right)\right]\mathrm{mol}}{6\mathrm{mol}+⌊\left(1/9\right)+\left(1/10\right)\mathrm{mol}\right]}=\frac{0.211\mathrm{mol}}{6.211\mathrm{mol}}=0.034$

Vapour pressure of solution. ${p}_{1}={x}_{2}{p}_{1}^{\ast }=\left(0.034\right)\left(35\mathrm{mmHg}\right)=1.19\mathrm{mmHg}$

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