Question

The vapour pressures of ethanol and methanol are $44.5\, mm \,Hg$ 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. The total vapour pressure of the solution and the mole fraction of methanol in the vapour are respectively

• A. $43.46\, mm$ and $0.51$
• B. $66.15\, mm$ and $0.657$
• C. $66.15 \,mm$ and $0.791$
• D. $70.59 \,mm$ and $0.657$

Answer 1

Answer: (B)

We know, $P_{TotaI} = p_1 + p_2$
Also $p_{1} = p^{\circ}_{1} \times x_{1} and p_{2} = p^{\circ}_{2} \times x_{2}$
$x_{1}$ (mole fraction of $CH_{3}OH$) $= \frac{\frac{40}{32}}{\frac{40}{32}+\frac{60}{46}} = 0.49$
[Mol. wt. of $CH_{3}OH = 32$, Mol. wt. of $C_{2}H_{5}OH = 46$]
$x_{2}$ (Mole fraction of $C_{2}H_{5}OH$) $= \frac{\frac{60}{40}}{\frac{40}{32}+\frac{60}{46}} = 0.51$
$p_{1} =$ Partial vapour pressure of $CH_{3}OH = 88.7 \times 0.49$
$= 43.46 \,mm\, Hg$
$p_{2} =$ Partial vapour pressure of $C_{2}H_{5}OH = 44.5 \times 0.51$
$= 22.69 \,mm \,Hg$
$P_{Total} =p_{1} +p_{2} = \left(43\cdot46 + 22\cdot69\right) = 66\cdot15\, mm\, Hg$
Mole fraction of $CH_{3}OH$ in vapour $= \frac{43.46}{66.15} = 0.657$