Vapour pressure of water at $293 \, K$ is $17.535 \, mm \, Hg$. Calculate the vapour pressure of water at $293 \, K$ when $25 \, g$ of glucose is dissolved in $450 \, g$ of water.

(N/A) Vapour pressure of pure water,$P_{1}^{\circ} = 17.535 \, mm \, Hg$
Mass of glucose,$w_{2} = 25 \, g$
Mass of water,$w_{1} = 450 \, g$
Molar mass of glucose $(C_{6}H_{12}O_{6})$,$M_{2} = 6 \times 12 + 12 \times 1 + 6 \times 16 = 180 \, g \, mol^{-1}$
Molar mass of water,$M_{1} = 18 \, g \, mol^{-1}$
Number of moles of glucose,$n_{2} = \frac{25 \, g}{180 \, g \, mol^{-1}} = 0.139 \, mol$
Number of moles of water,$n_{1} = \frac{450 \, g}{18 \, g \, mol^{-1}} = 25 \, mol$
Using Raoult's law for relative lowering of vapour pressure:
$\frac{P_{1}^{\circ} - p_{1}}{P_{1}^{\circ}} = \frac{n_{2}}{n_{2} + n_{1}}$
$\frac{17.535 - p_{1}}{17.535} = \frac{0.139}{0.139 + 25} = \frac{0.139}{25.139} \approx 0.00553$
$17.535 - p_{1} = 0.00553 \times 17.535 \approx 0.097$
$p_{1} = 17.535 - 0.097 = 17.438 \, mm \, Hg \approx 17.44 \, mm \, Hg$
The vapour pressure of the solution is $17.44 \, mm \, Hg$.