139.18 g of glucose is added to 178.2 g of water the vapour pressure of water for this aqueous solution at 100° C is

The molar masses of glucose and water are 180 g/mol and 18 g/mol respectively.

Number of moles of glucose = $\frac{139.18}{180}=0.773\approx 0.8$

Number of moles of water = $\frac{178.2}{18}=9.9$

Mole fraction of glucose =$\frac{0.8}{0.8+9.9}=0.075$

Relative lowering in vapour pressure is equal to the mole fraction of glucose.

$$\begin{gathered} \frac{{\mathrm{P}}^{\mathrm{o}}-\mathrm{P}}{{\mathrm{P}}^{\mathrm{o}}}={\mathrm{X}}_{\mathrm{s}}\left(\mathrm{glucose}\right) \\ \frac{760-\mathrm{P}}{760}=0.075 \\ \mathrm{P}=704\mathrm{torr}. \end{gathered}$$