Question

What is the partial vapor pressure of a benzene above solution obtained by mixing 74.9 g benzene, C6H6, and 33.9 g toluene, C6H5CH3, at 25 ∘C? At 25 ∘C the vapor pressure of C6H6 = 95.1 mmHg; the vapor pressure of C6H5CH3 = 28.4 mmHg.

What is the partial vapor pressure of a toluene?

What is the total vapor pressure of a solution?

Answer 1

1) Determine moles of benzene and toluene:

benzene ---> 74.9 g / 78.1134 g/mol = 0.95886 mol

toluene ---> 33.9 g / 92.1402 g/mol = 0.36791 mol

2) Determine the mole fraction of each substance:

benzene ---> 0.95886 mol / (0.95886 mol + 0.36791 mol) = 0.7227

toluene ---> 1 - 0.36791 = 0.63209

3) Raoult's Law for a solution of two volatiles is this:

P_total = P°_ben χ_ben + P°_tol χ_tol

P_total = (95.1) (0.7227) + (28.4) (0.63209)

P_total = 68.728 torr + 17.95 torr = 86.678 torr, round to 86.7 torr.