Methanol (p* = 90 mmHg at 300 K) and ethanol (p* = 51 mmHg at 300 K) form very nearly an ideal solution. The total vapour pressure of a solution obtained by mixing 23 g ethanol with 32 g methanol would be

Amount of methanol, $n_1=\frac{m_1}{M_1}=\frac{32\mathrm{g}}{32 \mathrm{g} {\mathrm{mol}}^{-1}}=1.0\mathrm{mol}$

Amount of ethanol, $n_2=\frac{m_2}{M_2}=\frac{23\mathrm{g}}{46 \mathrm{g} {\mathrm{mol}}^{-1}}=0.5 \mathrm{mol}$

Mole fraction of methanol, $x_1=\frac{n_1}{n_1+n_2}=\frac{1.0\mathrm{mol}}{(1.0+0.5)\mathrm{mol}}=2/3$

Total Vapour pressure, $p=x_1{p}_1^{\ast }+x_2{p}_2^{\ast }=\left[\right(2/3\left)\right(90)+(1/3\left)\right(51\left)\right]\mathrm{mmHg}=77\mathrm{mmH}g$