What weight of glucose must be dissolved in 100   g of water to lower the vapour pressure by 0 . 20…

Sol. Using the formula
$\frac{{\mathrm{P}}^{\mathrm{o}}-{\mathrm{P}}_{\mathrm{s}}}{{\mathrm{P}}^{\mathrm{o}}}=\frac{\mathrm{Moles} \mathrm{of} \mathrm{solute}}{\mathrm{Moles} \mathrm{of} \mathrm{solvent}}\left(\text{for dilute solution}\right)$

${\mathrm{P}}^{\mathrm{o}}=$Vapour pressure of pure water

${\mathrm{P}}_{\mathrm{s}}=$Vapour pressure of solution
$$\begin{gathered} \frac{54.2-54}{54.2}=\frac{\mathrm{w}/180}{{\displaystyle \frac{100}{18}}} \\ \mathrm{w}=\frac{0.2\times 1000}{54.2} \\ =\frac{200}{54.2}=3.69 \end{gathered}$$