The vapour pressure of pure water at $302 K$ is $31.80 mmHg$ . Find the lowering of vapour pressure, in mmHg, when $1.8 g$ of glucose (molar mass $180$ ) is dissolved in $36 g$ of water (molar mass $18$ ). Give the answer to three decimal places.

Numerical answer type, answer in mmHg, tolerance 0.001.

Answer: 0.158 mmHg

n_{2} = 1.8/180 = 0.0100

and

n_{1} = 36/18 = 2.00

Δ p = p^{0} x_{2} = 31.80 \times 0.0100/2.0100 = 0.1582 mmHg