Question

The vapour pressure of ethanol and methanol are $42.0\text{mm}$ and $88.5\text{mm Hg}$ respectively. An ideal solution is formed at the same temperature by mixing $46.0\text{g}$ of ethanol with $16.0\text{g}$ of methanol. The mole fraction of methanol in the vapour is:

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

Answer 1

The correct option is C $0.513$

According to Raoult’s law;

$P_{total}=P_1+P_2$

$P_1$ = Partial pressure of one component
$P_2$ = Partial pressure of other component
$Moles\left(n\right)=\frac{Givenmass}{Molarmass}$

$n_{C{H}_{3}OH}=\frac{16}{32}=\frac{1}{2}\text{mol}$ (Molar mass of $C{H}_{3}OH=32\text{g/mol})$

$n_{C_2{H}_{5}OH}=\frac{46}{46}=1\text{mol}$ (Molar mass of $C_2{H}_{5}OH=46\text{g/mol})$

$X_{C_2{H}_{5}OH}\left(molefraction\right)=\frac{1}{1+\frac{1}{2}}=\frac{2}{3}\left\{\therefore X_A+X_B=1\right\}{X}_{C{H}_{3}OH}=1-\frac{2}{3}=\frac{1}{3}$

As we know $P_1=X_1{p}_1^0$

So, Partial vapour pressure of $C{H}_{3}OH\left(P_1\right)=88.5\times \frac{1}{3}=29.5mmHg$

Partial vapour pressure of $C_2{H}_{5}OH\left(P_2\right)=42\times \frac{2}{3}=28\text{mm Hg}$

$\text{Total partial pressure = 29.5 + 28 = 57.5}$

Mole fraction of methanol in vapour $=\frac{29.5}{57.5}=0.513$

Therefore, the correct answer is (C).