What mass of glucose (molar mass $180 g mo l^{- 1}$ ) must be dissolved in $250 g$ of water (molar mass $18 g mo l^{- 1}$ ) to lower the vapour pressure of water by $2.0%$ of its pure value? Use the exact mole-fraction form of Raoult's law and give the answer in grams to the nearest integer.

Numerical answer type, answer in g, tolerance 1.

Answer: 51 g

\frac{p ^{0} - p}{p ^{0}} = x_{2} = 0.020

, so

\frac{n _{2}}{n _{2} + n _{1}} = 0.020

and

n_{2} = \frac{0.020}{0.980} n_{1}

. With

n_{1} = 250/18 = 13.889

n_{2} = 0.2834 mol

. Mass

= 0.2834 \times 180 = 51.02 \approx 51 g