A sample of 20.0 g of a compound ( molecular weight 120) which is a non-electrolyte is dissolved in 10.0 grams of ethanol $(C_{2} H_{5} OH)$ . If the vapour pressure of pure ethanol at the temperature used is 0.250 atm, what is the vapour pressure of the solution?

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0.83 atm

0.125 atm

0.250 atm

0.141 atm

answer is D.

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Detailed Solution

$\frac{P^{o} - P_{s}}{P^{o}} = \frac{n}{n + N}$

$\frac{0 .25 - P_{s}}{0 .25} = \frac{\frac{20}{120}}{\frac{20}{120} + \frac{10}{46}} = \frac{0 .167}{0 .167 + 0 .217}$

$∴ 0 .25 - P_{s} = \frac{0 .25 \times 0 .167}{0 .167 + 0 .217} ∴ P_{s} = 0 .141$

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