Question

At a certain temperature the vapor pressure of pure benzene is measured to be 322. torr . Suppose a solution is prepared by mixing 138.g of benzene and 91.2g of heptane .

Calculate the partial pressure of benzene vapor above this solution. 3 sig figs!
Steps/equations are appreciated...!

Answer 1

First, we convert the masses into fractions
138 g of benzene = 1.77 moles benzene
91.2 g of heptane = 0.93 moles heptane

Next, calculate the mole fractions
x1 (benzene) = 1.77 / (1.77 + 0.93) = 0.66
x2 (heptane) = 1 - 0.66 = 0.34

The vapor pressure of pure benzene is 322 torr at 40 C.
At the same temperature, the vapor pressure of heptane is 92 torr.

The partial pressure of benzene is
p1 = 0.66 (322) = 212.52 torr
The partial pressure of heptane is
p2 = 0.34 (92) = 31.28 torr