The vapour pressure lowering caused by the addition of 100 g 

of sucrose(molecular mass = 342) to 1000 g of water if the vapour

 pressure of pure water at 25 ° C is 23.8 mm Hg

Given molecular mass of sucrose =  342

Moles of sucrose = $=\frac{100}{342}=0.292$  mole

Moles of water   $N=\frac{1000}{18}=55.5$  moles and

Vapour pressure of pure water P 0 = 23.8 mm Hg

According to Raoult's law  

 $\begin{array}{l}\frac{\mathrm{\Delta }P}{P^{\circ }}=\frac{n}{n+N}\Rightarrow \frac{\mathrm{\Delta }P}{23.8}=\frac{0.292}{0.292+55.5}\\ \mathrm{\Delta }P=\frac{23.8\times 0.292}{55.792}=0.125\mathrm{mmHg}\end{array}$