The vapour pressure of pure water at $300 K$ is $27.0$ torr. Find the vapour pressure (torr, two decimal places) of a solution of $36 g$ glucose (molar mass $180 g mo l^{- 1}$ ) in $360 g$ water (molar mass $18 g mo l^{- 1}$ ), using Raoult's law.

Numerical answer type, answer in torr, tolerance 0.01.

Answer: 26.73 torr

n_{2} = 0.200

mol and

n_{1} = 20.0

mol.

x_{1} = 20.0/20.2 = 0.9901

P = 27.0 \times 0.9901 = 26.73

torr.