Biphenyl, C12H10, is a nonvolatile, nonionizing solute that is soluble in benzene, C6H6. At 25 °C, the vapor pressure of pure benzene is 100.84 torr. What is the vapor pressure of a solution made from dissolving 19.6 g of biphenyl in 25.1 g of benzene?

3 answers

mols biphenyl = grams/molar mass.
mols benzene = grams/molar mass.
Xbenzene = nbenzene/total mols.

Psoln = Xbenzene*Pobenzene
19.6 g of biphenyl /154.21 g of biphenyl*mol-1= moles of biphenyl

25.1 g of benzene/78.11 g of benzene *mol-1= moles of benzene

Since at 25°C, the vapor pressure of pure benzene is 100.84 torr, then the pressure must be equal to the mole fraction of benzene to the total number of moles in the solution

100.84 torr*(moles of benzene/[moles of benzene + moles of biphenyl])= pressure of the solution

Dr. Bob222 posted the solution to a problem like this in past posts
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