Total vapour pressure (P) of an ideal solution made up of ethanol and methanol is given by, $$P = {x_1}p_1^0 + {x_2}p_2^0$$ where $$p_1^0$$ & $$p_2^0$$ are the vapour pressures of pure ethanol and pure methanol respectively and x₁ and x₂ are the mole fractions of ethanol and methanol respectively in solution.

Now, number of moles of ethanol $$ = {{60} \over {46}} = 1.3$$ (Molar mass of ethanol = 46 g mol^$$-$$1

and that of methanol $$ = {{40} \over {32}} = 1.25$$ (Molar mass of methanol = 32 g mol^$$-$$1

Total moles of ethanol & methanol = (1.3 + 1.25) = 2.55 mol

$$\therefore$$ $${x_1} = {{1.3} \over {2.55}} = 0.50$$ and $${x_2} = {{1.25} \over {2.55}} = 0.49$$

Given : $$p_1^0$$ = 44.5 mm Hg and $$p_2^0$$ = 88.7 mm Hg

$$\therefore$$ $$p = {x_1}p_1^0 + {x_2}p_2^0 = (0.50 \times 44.5 + 0.49 \times 88.7)$$ mm Hg = 65.71 mm Hg

$$\therefore$$ Total vapour pressure of the solution = 65.71 mm Hg

mole fraction of methanol in vapour $$ = {{Partial\,v.p.\,of\,methanol\,in\,vapour\,of\,the\,solution} \over {Total\,v.p.\,of\,solution}}$$

$$ = {{{x_2}p_2^0} \over P} = {{0.49 \times 88.7} \over {65.71}} = 0.66$$