Question

the vapour pressure lowering caused by the addition of 100g of sucrose(molecular mass=342) to 1000g of water having the vapour pressure of pure water at 25^*C is 23.8 mm Hg.

Answer 1

$$\begin{gathered} \text{Dear Student,} \\ \mathrm{According}\mathrm{to}\mathrm{Raoult}\text{'}\mathrm{s}\mathrm{law}, \\ \frac{\mathrm{P}-\mathrm{Ps}}{\mathrm{P}}={\mathrm{X}}_{\mathrm{solute}}=\frac{\mathrm{No}.\mathrm{of}\mathrm{moles}\mathrm{of}\mathrm{solute}\left(\mathrm{n}\right)}{\mathrm{Total}\mathrm{no}.\mathrm{of}\mathrm{solute}}=\frac{\mathrm{n}}{\mathrm{n}+\mathrm{N}} \\ \frac{∆\mathrm{P}}{\mathrm{P}}=\frac{\mathrm{n}}{\mathrm{n}+\mathrm{N}} \\ \mathrm{n}=\frac{100}{342}=0.292 \\ \mathrm{N}=\frac{1000}{18}=55.5 \\ \mathrm{P}=23.8\mathrm{mm}\mathrm{Hg} \\ \frac{∆\mathrm{P}}{23.8}=\frac{0.292}{0.292+55.5} \\ ∆\mathrm{P}=0.125\mathrm{mm}\mathrm{Hg} \end{gathered}$$