18 g of glucose (C₆ H₁₂O₆ ) is added to 178.2 g water. The vapour pressure (in torr) for this aqueous solution is

Vapour pressure of water $\left({\mathrm{p}}^{\circ }\right)=760 \mathrm{torr}$
Number of moles of glucose
 $\begin{array}{l}=\frac{\text{ Mass of glucose (in g) }}{\text{ Molecular mass of glucose }\left(\mathrm{g}{\mathrm{mol}}^{-1}\right)}\\ =\frac{18\mathrm{g}}{180\mathrm{g}{\mathrm{mol}}^{-1}}=0.1\mathrm{mol}\end{array}$
Number of moles of water $=\frac{178.2\mathrm{g}}{18\mathrm{g}{\mathrm{mol}}^{-1}}=9.9\mathrm{mol}$
Total number of moles $=(0.1+9.9)\text{ moles }=10\mathrm{mol}$
Now, mole fraction of glucose in solution = Change in pressure with respect to initial pressure
                                     $\frac{\mathrm{\Delta p}}{{\mathrm{p}}^{\circ }}={\mathrm{X}}_{\mathrm{s}}=\frac{{\mathrm{n}}_{\mathrm{s}}}{{\mathrm{n}}_{\mathrm{s}}+{\mathrm{n}}_{\mathrm{o}}}=\frac{0.1}{10}$
or                         $\mathrm{\Delta p}=0.01{\mathrm{p}}^{\circ }=0.01\times 760=7.6 \mathrm{torr}$
$\because \mathrm{Vapour} \mathrm{pressure} \mathrm{of} \mathrm{solution} =(760-7.6)=752.4 \mathrm{torr}.$