At a certain temperature the vapour pressure of pure water is $760.0$ mmHg. A solution of $18.0$ g of a non-volatile non-electrolyte in $90.0$ g of water ( $18 g mo l^{- 1}$ ) has a vapour pressure of $745.1$ mmHg. Use $\frac{P ^{0} - P}{P} = \frac{n _{2}}{n _{1}}$ . Find the molar mass of the solute in $g mo l^{- 1}$ (nearest integer).

Numerical answer type, answer in g/mol, tolerance 1.

Answer: 180 g/mol

n_{1} = 90.0/18 = 5.00

mol.

(P^{0} - P) / P = (760.0 - 745.1) /745.1 = 0.02000

, so

n_{2} = 0.02000 \times 5.00 = 0.1000

mol.

M = 18.0/0.1000 = 180.0 g mo l^{- 1}

, i.e. about 180.