Question

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 11.5 g of biphenyl in 31.9 g of benzene?

Answer 1

Answer:

$P_{solution} = 85.3Torr$

Explanation:

Raoult's law is a tool that allows to determine vapour pressure of solutions. The formula is:

$P_{solution} = X_{solvent}P_{solvent}$ (1)

Where

P is Pressure of solution and solvent and X is mole fraction.

Moles of solute and solvent are:

Biphenyl:

11.5g×(1mol /154.21g) = 0.0746mol

Benzene

31.9g×(1mol /78.11g) = 0.408mol

Mole fraction of benzene is:

$\frac{0.408mol}{0.408mol + 0.0746mol}$ = 0.846

Replacing in (1):

$P_{solution} = 0.846*100.84Torr$

$P_{solution} = 85.3Torr$

I hope it helps!