Molar mass

Chemistry Level 2

The vapour pressure of acetone at ${20}^\circ$ $C$ is $185$ torr. When $1.2$ $g$ of a non-volatile substance was dissolved in $100$ $g$ of acetone at ${20}^\circ$ $C,$ its vapour pressure was $183$ torr. What is the molar mass $(g$ $mo1^{-1})$ of the substance $?$

The answer is 64.

1 solution

Marc Fernández
Jul 23, 2017

P* is vapor preasure of acetone, P is vapor preasure of solution. 1 is for acetone and 2 for non-volatile substance

$P = P^{*}\times \chi_{1} = P^{*} \frac{n_{1}}{n_{1}+n_{2}} = P^{*} \frac{m_{1}/M{1}}{m_{1}/M{1}+m_{2}/M{2}}$

$M_{2} = \frac{m_{2}}{\frac{m_{1}}{M_{1}}(\frac{P^{*}}{P}-1)} = 64 \ g/mol$

Molar mass

The answer is 64.

1 solution

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