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 gm of ethanol with 16.0 gm of methanol. The mole fraction of methanol vapour is:

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

Answer 1

The correct option is C $0.513$

$X_{C{H}_{3}OH}=\frac{n_{C{H}_{3}OH}}{n_{C{H}_{3}OH}+n_{C_2{H}_{3}OH}}$
$\frac{\frac{16}{32}}{\frac{16}{32}+\frac{46}{46}}=\frac{0\cdot 5}{1\cdot 5}$
$X_{C{H}_{3}OH}=\frac{1}{3},$

Similarly, $X_{C_2{H}_{5}OH}=\frac{2}{3}$

$P_{total}=X_A{P}_A+X_B{P}_B=42\times \frac{2}{3}+88\cdot 5\times \frac{1}{3}=57.5$
So, the mole fraction of methanol vapor $=\frac{P_B\times X_B}{P_{total}}=\frac{88.5\times \frac{1}{3}}{57\cdot 5}$$=0\cdot 513$

Hence, the correct option is $C$