Vapour pressure of dilute aqueous solution of glucose is 750 mm of Hg at 373 K. What is the mole fraction of solute if Vapour pressure of pure solvent is 760 mm Hg?

\(\frac{1}{76}\) \(\frac{1}{7.6}\) \(\frac{1}{38}\) \(\frac{1}{10}\)

\(\frac{1}{76}\)

\(\frac{Po - P}{Po}\) = X2 \(\frac{760 - 750}{760}\) = X2 or X2 = \(\frac{1}{76}\)