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 ... !

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