The amount of solution (molar mass,$60 \mathrm{g} {\mathrm{mol}}^{-}$) that must be added to $180 \mathrm{g}$ water so that the vapour pressure of water is lowered by $10 \%$ is

$$\begin{gathered} \mathrm{Relative} \mathrm{lowering} \mathrm{of} \mathrm{vapour} \mathrm{pressure}, \frac{{\mathrm{p}}^{\mathrm{o}}-{\mathrm{p}}_{\mathrm{s}}}{{\mathrm{p}}^{\mathrm{o}}}={\mathrm{X}}_{\mathrm{solute}} \\ {\mathrm{p}}^{\mathrm{o}}=\mathrm{vapour} \mathrm{pressure} \mathrm{of} \mathrm{the} \mathrm{pure} \mathrm{solvent} \\ {\mathrm{p}}_{\mathrm{s}}=\mathrm{vapour} \mathrm{pressure} \mathrm{of} \mathrm{the} \mathrm{solution} \\ {\mathrm{X}}_{\mathrm{solute}}=\mathrm{mole} \mathrm{fraction} \mathrm{of} \mathrm{the} \mathrm{solute} \\ \frac{100-90}{100}=\frac{{\displaystyle \frac{\mathrm{x}}{60}}}{{\displaystyle \frac{180}{18}}} \\ \frac{10}{100}=\frac{\mathrm{x}}{60}\times \frac{18}{180} \\ \mathrm{x}=60 \mathrm{g} \end{gathered}$$