Question

The vapour pressures of ethanol and methanol are 44.5 mm and 88.7 mm Hg respectively. An ideal solution is formed at the same temperature by mixing 60 g of ethanol with 40 g of methanol. Calculate the total vapour pressure of the solution and the mole fraction of methanol in the vapour.

Answer 1

Mol. mass of ethyl alcohol $=C_2{H}_{5}OH=46$

No. of moles of ethyl alcohol $=\frac{60}{46}=1.304$

Mol. mass of methyl alcohol $=C{H}_{3}OH=32$

No. of moles of methyl alcohol $=\frac{40}{32}=1.25$

${}^{\prime }{X}_A^{\prime }$, mole fraction of ethyl alcohol $=\frac{1.304}{1.304+1.25}=0.5107$

${}^{\prime }{X}_B^{\prime }$, mole fraction of methyl alcohol $=\frac{1.25}{1.304+1.25}=0.4893$

Partial pressure ethyl alcohol $=X_A{\dot{p}}_A^0=0.5107\times 44.5=22.73mmHg$

Partial pressure methyl alcohol $=X_B{\dot{p}}_B^0=0.4893\times 88.7=73.40mmHg$

Total vapour pressure of solution $=22.73+43.40=66.13mmHg$

Mole fraction of methyl alcohol in the vapour

$=\frac{{\text{Partial pressure of CH}}_{3}OH}{\text{Total vapour pressure}}=\frac{43.40}{66.13}=0.6563$