Question

The vapour pressure of ethanol and methanol are $44.5mmHg$ and $88.7mmHg$, respectively. An ideal solution is fromed at the same temperature by mixing $60g$ of ethanol and $40g$ of methanol. Calculate the total vapour pressure of the solution and the mole fraction of methanol in the vapour.

• A. $33.55mmHg,0.74$
• B. $66.11mmHg,0.656$
• C. $92.87mmHg,0.725$
• D. $132.22mmHg,0.945$

Answer 1

The correct option is B $66.11mmHg,0.656$

The molar masses of ethanol and methanol are $46$ g/mol and $32$ g/mol respectively.

$60$ g ethanol $=\frac{60}{46}=1.3$ moles.

$40$ g methanol $=\frac{40}{32}=1.25$ moles.

Mole fraction of ethanol $=\frac{1.3}{1.3+1.25}=0.51$

Mole fraction of methanol $=1-0.51=0.49$

The total vapour pressure of the solution $=P_{ethanol}^0{X}_{ethanol}+P_{methanol}^0{X}_{methanol}$

$=44.5\times 0.51+88.7\times 0.49=66.11$ mm Hg.

The mole fraction of methanol in the vapour $\frac{P_{methanol}}{P_{total}}=\frac{88.7\times 0.49}{66.11}=0.656$