Question

The vapour pressure of a mixture of diethyl aniline and water is $1.013\times {10}^{5}N{m}^{-2}$ at ${99.4}^{o}C$. The vapour pressure of water at this temperature is $0.992\times {10}^{5}N{m}^{-2}$. The grams of steam are necessary to distill over 100 g of diethylaniline are (write answer as nearest integer after dividing by 60).

Answer 1

Let w g of steam be required.The molar mass of diethylaniline is $149.23$ g/mol. The number of moles of diethylaniline in $100$ g be $\frac{100}{149.23}=0.67$ moles.
But $\frac{p^0-p}{p^0}=\frac{n}{n+N}$
$\frac{1.013-0.992}{0.992}=\frac{0.67}{0.67+\frac{w}{18}}$
$0.0212=\frac{1}{1+\frac{w}{12.06}}$
$w+12.06=569.7$
$w=557.6$
Hence, $557.6$ grams of steam are necessary to distill over $100$ g of diethylaniline.