Question

Biphenyl, C12H10, is a nonvolatile, nonionizing solute that is soluble in benzene, C6H6. at 25 degrees Celsius, the vapor pressure of pure benzene is 100.84 torr. what is the vapor pressure of a solution made from dissolving 20.0 g of biphenyl in 32.6 g of benzene?

Answer 1

At first we need to find the moles of biphenyl, C₁₂H₁₀ as---

moles C₁₂H₁₀ =( 20.0 g C₁₂H₁₀ / 154.21 g C₁₂H₁₀ mol^-1) = 0.12969 mol C₁₂H₁₀

and moles of benzene, C₆H₆ can be calculated as--

moles C₆H₆ =(32.6 g C₆H₆ / 78.11 g C₆H₆ mol^-1) = 0.41736 mol  C₆H₆

Now, the vapor pressure of the solution will be--

(Vapor pressure of pure benzene* mol of benzene) /(mol of biphenyl + mol of benzene)

=(100.84 torr) * (0.41736 mol) / (0.12969 mol + 0.41736 mol)

=(42.086 / 0.54705) torr

=76.93 torr