The vapour pressure of acetone at 20 °C is 185 Torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20 °C, its vapour pressure was 193 Torr. The molar mass of the substance is

# The vapour pressure of acetone at 20 °C is 185 Torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20 °C, its vapour pressure was 193 Torr. The molar mass of the substance is

1. A

$32\mathrm{g}{\mathrm{mol}}^{-1}$

2. B

$64\mathrm{g}{\mathrm{mol}}^{-1}$

3. C

$128\mathrm{g}{\mathrm{mol}}^{-1}$

4. D

$488\mathrm{g}{\mathrm{mol}}^{-1}$

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### Solution:

According to Raoult's law,

Also,    ${x}_{2}=\frac{{n}_{2}}{{n}_{1}+{n}_{2}}=\frac{\left(1.2\mathrm{g}/M\right)}{\left(100\mathrm{g}/58\mathrm{g}{\mathrm{mol}}^{-1}\right)+\left(1.2\mathrm{g}/M\right)}$

Hence,   $\frac{\left(1.2\mathrm{g}/M\right)}{\left(100\mathrm{g}/58\mathrm{g}{\mathrm{mol}}^{-1}\right)+\left(1.2\mathrm{g}/M\right)}=\frac{2}{185}$

This gives

$\frac{1.2\mathrm{g}}{M}\left(1-\frac{2}{185}\right)=\left(\frac{2}{185}\right)\left(\frac{100\mathrm{g}}{58\mathrm{g}{\mathrm{mol}}^{-1}}\right)$  or  $M=\left(1.2\mathrm{g}\right)\left(\frac{183}{185}\right)\left(\frac{185}{2}\right)\left(\frac{58\mathrm{g}{\mathrm{mol}}^{-1}}{100\mathrm{g}}\right)=63.68\mathrm{g}{\mathrm{mol}}^{-1}$

Approixmate value Writing

This gives $M=\left(1.2×\frac{58}{100}×\frac{185}{2}\right)\mathrm{g}{\mathrm{mol}}^{-1}=64.38\mathrm{g}{\mathrm{mol}}^{-1}$

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