Question

At ${25}^{o}C$, the vapour pressure of pure benzene is 100 torr, while that of pure ethyl alcohol is 44 torr. Assuming ideal behaviour, calculate the vapour pressure at ${25}^{o}C$ of a solution which contains 10 g of each substance.

• A. 33.775 torr
• B. 54.775 torr
• C. 64.775 torr
• D. 60.775 torr

Answer 1

The correct option is C 64.775 torr

We have,

moles of benzene = $\frac{10}{78}$ = 0.128

moles of ethyl alcohol = $\frac{10}{46}$ = 0.217

$P_t=P_A^0\ast x_A+P_B^0\ast x_B$

$P_t=100\ast \frac{0.128}{0.345}+44\ast \frac{0.217}{.345}$

$P_t=64.77torr$