Question

The vapour pressures of ethanol and methanol are 42.0 mm $Hg$ and 88.5 mm $Hg$ respectively. An ideal solution is formed at the same temperature by mixing 46.0 g of ethanol with 16.0 g of methanol. What is the mole fraction of methanol vapour ?

• A. 0.467
• B. 0.502
• C. 0.513
• D. 0.556

Answer 1

The correct option is C 0.513

$\begin{array}{ccc}& C{H}_{3}OH\left(A\right)& C_2{H}_{5}OH\left(B\right)\\ \text{vapour pressure}& 88.5\text{mm}Hg& 42.0\text{mm}Hg\\ \text{mass}& 16.0\text{g}& 46.0\text{g}\\ \text{moles}& 0.5& 1\end{array}$
$\therefore$ Total moles = 1 + 0.5 = 1.5
$P_{\text{total}}=P_A^{\circ }{X}_A+P_B^{\circ }{X}_B$
$=88.5\times \frac{0.5}{1.5}+42\times \frac{1}{1.5}$
$P_{\text{total}}=29.5+28=57.5\text{mm}Hg$
Mole fraction of methanol in vapour phase is $Y_A$.
$P_A=Y_A\times \text{total pressure}$
$P_A=P_A^{\circ }{X}_A=29.5$
$Y_A=\frac{29.5}{57.5}=0.513$