Question

The vapour pressure of ethanol (A) and methanol (B) are $44.5mmHg$ and $88.7mmHg$ respectively. An ideal solution is formed at the same temperature by mixing $60g$ of ethanol with $40g$ of methanol. Calculate the total vapour pressure of the solution $\left(P_T\right)$ and the mole fraction of methanol in the vapour phase $\left(Y_B\right)$

• A. $P_T=88.33mmHg$ and $Y_B=0.25$
• B. $P_T=66.13mmHg$ and $Y_B=0.25$
• C. $P_T=88.33mmHg$ and $Y_B=0.65$
• D. $P_T=66.13mmHg$ and $Y_B=0.65$

Answer 1

The correct option is D $P_T=66.13mmHg$ and $Y_B=0.65$

$\text{Molar mass of ethyl alcohol}=46gmo{l}^{-1}$
$\text{No. of moles of ethyl alcohol}=\frac{60}{46}=1.304mol$
$\text{Molar mass of methyl alcohol}=32gmo{l}^{-1}$
$\text{No. of moles of methyl alcohol}=\frac{40}{32}=1.25mol$
$\text{Mole fraction of ethyl alcohol,}{X}_A=\frac{1.304}{1.304+1.25}=0.5107$
$\text{Mole fraction of methyl alcohol,}{X}_B=\frac{1.25}{1.304+1.25}=0.4893$
$\text{Partial pressure of ethyl alcohol}=X_A\times P_A^{\circ }=0.5107\times 44.5=22.73mmHg$
$\text{Partial pressure of methyl alcohol}=X_B\times P_B^{\circ }=0.4893\times 88.7=43.40mmHg$
$\text{Total vapour pressure of solution}=22.73+43.40=66.13mmHg$
$\text{Mole fraction of methyl alcohol in the vapour,}{Y}_B=\frac{\text{Partial pressure of}C{H}_{3}OH}{\text{Total vapour pressure}}=\frac{43.40}{66.13}=0.6563$