An ideal solution was prepared by mixing $40 \mathrm{g}$ of methyl alcohol and $60 \mathrm{g}$ of ethyl alcohol. If vapour pressure of methyl alcohol and ethyl alcohol respectively are    $88.7 \mathrm{and} 44.5 \mathrm{mm} \mathrm{Hg}$, then find the total pressure of the solution

  Number of moles of  ${\mathrm{CH}}_3\mathrm{OH}=\frac{\mathrm{wt}}{\mathrm{mol}\cdot \mathrm{wt}}=\frac{40 \mathrm{g}}{32 \mathrm{g} {\mathrm{mol}}^{-1}}=1.25$

  Number of moles of $\begin{array}{c}{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}=\frac{\mathrm{wt}}{\mathrm{mol}.\mathrm{wt}}=\frac{60 \mathrm{g}}{46 \mathrm{g} {\mathrm{mol}}^{-1}}=1.3 \mathrm{mol}\\ \end{array}$

$x_A$ = (mole fraction of  ${\mathrm{CH}}_3\mathrm{OH}$ ) $=\frac{1.25}{1.25+1.30}=\frac{1.25}{2.55}=0.49$

$x_B$ = (mole fraction of ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}=1-0.49=0.51$ 

Total pressure of solution 

             $\begin{array}{l}P=x_A{P_A}^0+x_B{P_B}^0\\ =\left(0.49\times 88.7\right)+\left(0.51\times 44.5\right)\\ =43.46+22.69\\ =66.15 \mathrm{mm} \mathrm{Hg}\end{array}$