Given, at 373 K vapour pressure of the aqueous solution of glucose, p = 750 mm Hg.

We know, boiling point of water = 373 K. At this temperature, vapour pressure of water, p0
= 760 mm Hg

According to Raoults' law, $${{{p^0} - p} \over {{p^0}}} = {x_2} \approx {{{n_2}} \over {{n_1}}}$$

$$\therefore$$ $${{760 - 750} \over {760}} = {{{n_2}} \over {{n_1}}}$$ or, $${n_2} = {n_1} \times 0.013$$

(1) Amount of water in solution n₁ mol = 18 $$\times$$ n₁g

$$\therefore$$ Molality of the solution $$ = {{{n_2} \times 1000} \over {18 \times {n_1}}} = {{{n_1} \times 0.013 \times 1000} \over {18 \times {n_1}}}$$ = 0.72 mol kg^$$-$$1

(2) Mole fraction of solute in solution $$ = {{{n_2}} \over {{n_1} + {n_2}}} = {{{n_1} \times 0.013} \over {{n_1} + {n_1} \times 0.013}} = 0.0128$$