JEE Main 2015 · 4 Apr · Shift 1 · Chemistry Q36

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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 183 torr. The molar mass (g mol $^{-1}$ ) of the substance is :

128

488

Answer and worked solution

Official answer: B

$\dfrac{\Delta p}{p^\circ}=\dfrac{2}{185}=\dfrac{n_2}{n_1}$ (dilute solution), with $n_1=\frac{100}{58}=1.724$ mol of acetone. So $n_2=0.01864$ mol and
$M=\frac{1.2}{0.01864}\approx64\text{ g mol}^{-1}.$

Topics:Colligative properties Raoult's law and vapour pressure

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JEE Main 2015 · 4 Apr · Shift 1, Chemistry question 36