Name: Raoult's Law
Uploaded: 2019-07-19
Duration: 3 min 24 s
Description: Raoult's law is a chemical law that states that the vapour pressure of a solution is dependent on the mole fraction of a solute added to the solution. Learn...

Question

Ethanol and methanol form a solution that is very nearly ideal. The vapour pressure of ethanol is $44.5 mm Hg$ and that of methanol is $88.7 mm Hg$ at $20 ° C$ .

Calculate the partial pressure and the total vapour pressure of this solution obtained by mixing $60 g$ of ethanol with $40 g$ of methanol, and the mole fraction of ethanol in the vapour.

Accepted Answer

Given, weight of methanol = $40 g$

Moles of methanol, $(n_{methanol})$ = $\frac{Weight of methanol ({CH}_{3} OH)}{Molar mass of methanol} = \frac{40 g}{32 g {mol}^{‐ 1}} = 1.25 mol$

Weight of ethanol = $60 g$

Moles of ethanol, $(n_{ethanol})$ = $\frac{Weight of ethanol (C_{2} H_{5} OH)}{Molar mass of ethanol} = \frac{60 g}{46 g {mol}^{‐ 1}} = 1.30 mol$

Mole fraction of methanol, $(x_{methanol}) = \frac{Moles of methanol}{Moles of methanol + Moles of ethanol} = \frac{1.25 mol}{1.25 mol + 1.30 mol} = 0.49$

Moles fraction of ethanol, $(x_{ethanol}) = \frac{Moles of ethanol}{Moles of ethanol + Moles of methanol} = \frac{1.30 mol}{1.30 mol + 1.25 mol} = 0.51$

So, moles of methanol and ethanol in the solution is $0.49$ mol and $0.51$ mol, respectively.

As given, $P_{methanol}^{\circ} = 88.7 mm of Hg, P_{ethanol}^{\circ} = 44.5 mm of Hg$

So Total vapour pressure, $P_{Total} = P_{methanol}^{\circ} . x_{methanol} + P_{ethanol}^{\circ} . x_{ethanol} = (88.7 \times 0.49 + 44.5 \times 0.51) mm of Hg = 66.16 mm of Hg$

Partial pressure of methanol, $P_{methanol}^{'} = P_{methanol}^{\circ} . x_{methanol} = (88.7 \times 0.49) mm of Hg = 43.46 mm of Hg$

Partial pressure of ethanol, $P_{ethanol}^{'} = P_{ethanol}^{\circ} . x_{ethanol} = (44.5 \times 0.51) mm of Hg = 22.7 mm of Hg$

Mole fraction of methanol in vapour phase, $x_{methanol}^{v} = \frac{Partial pressure of methanol}{Total vapour pressure} = \frac{43.46}{66.16} = 0.66$

Mole fraction of ethanol in vapour phase, $x_{ethanol}^{v} = \frac{Partial pressure of ethanol}{Total vapour pressure} = \frac{22.7}{66.16} = 0.34$

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