At $10 0^{\circ} C$ the vapour pressure of pure water is $760 mmHg$ . When $6.0 g$ of a non-volatile non-electrolyte is dissolved in $90 g$ of water, the vapour pressure falls by $7.5 mmHg$ . Use the exact form of Raoult's law with mole fractions (molar mass of water $18 g mo l^{- 1}$ ). Find the molar mass of the solute in $g mo l^{- 1}$ , to the nearest integer.

Numerical answer type, answer in g/mol, tolerance 2.

Answer: 120 g/mol

\frac{Δ p}{p ^{0}} = x_{2} = \frac{n _{2}}{n _{2} + n _{1}}

with

n_{1} = 90/18 = 5.00

. Here

x_{2} = 7.5/760 = 0.00987

, so

n_{2} = x_{2} n_{1} / (1 - x_{2}) = 0.04983 mol

. Molar mass

= 6.0/0.04983 = 120.4 \approx 120 g mo l^{- 1}