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Raoult's Law

The vapour pr...

Question

The vapour pressure of ethanol and methanol are 44.5 mm and 88.7 mm Hg respectively. An ideal solution is prepared at the same temperature by mixing 60 g of ethanol with 40 g of methanol. The total vapour pressure of the solution is:

A

67.13 mm Hg

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B

66.13 mm Hg

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C

76.13 mm Hg

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D

None of the above

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Solution

The correct option is C 66.13 mm Hg
Ethanol
$C H_{3} C H_{2} O H$
Mol. weight = 42 gms
Mass added in the solution = 60 gms
$M o l e_{C H_{3} C H_{2} O H} = 60 / 42 = 10 / 7$

Methanol
$C H_{3} O H$
Mol. weight = 28 gms
Mass added in the solution = 40 gms
$M o l e_{C H_{3} O H} = 40 / 28 = 10 / 7$

Mole Ratio
$n_{C H_{3} C H_{2} O H} = \frac{M o l e_{C H_{3} C H_{2} O H}}{N_{t o t a l}} = 0.5$
$n_{C H_{3} O H} = \frac{M o l e_{C H_{3} O H}}{N_{t o t a l}} = 0.5$

Vapour Pressure Calculations

Vapor Pressure of Solution = $n_{C H_{3} C H_{2} O H} \times P_{C H_{3} C H_{2} O H}^{v a p} + n_{C H_{3} O H} \times P_{C H_{3} O H}^{v a p}$
= $0.5 \times 44.5 + 0.5 \times 88.7$ mm
= 66.13 mm

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