Find the value of mole fraction of the solute when vapour pressure of glucose is 750 mm Hg at 373 K in dilute solution.

$\begin{array}{l}\frac{{\mathrm{P}}^{\circ }-{\mathrm{P}}_{\mathrm{s}}}{{\mathrm{P}}^{\circ }}=\frac{\mathrm{n}}{\mathrm{n}+\mathrm{N}}\\ \therefore \frac{\mathrm{n}}{\mathrm{n}+\mathrm{N}}=\frac{760-750}{760}=\frac{1}{76}\end{array}$

n = number of moles of glucose 

N = number of moles of water

At boiling point of water P⁰ = 1 atm = 760 mm Hg