Question

1) Calculate the mole fraction of benzene and toluene in the vapor in a solution of both with Xbenzene = 0.763. (At 25°C, the vapor pressure of benzene is 96.0 mm Hg and 30.3 mm Hg for toluene).

2) Calculate the freezing point of a solution prepared by dissolving 15.0 g of Na2SO4 in 100 g of water. Kf for water is 1.86 °C.kg/mol.

Answer 1

total P = 0.763 x 96.0 mm Hg + 0.237 x 30.3 mm Hg

= 73.25 mm Hg + 7.18 mm Hg

= 80.43 mm of Hg

Mole fraction of benzene in vapor phase = partial pressure of benzene / total pressure

=73.25 / 80.43 = 0.91

Mole fraction of toluene in vapor phase = partial pressure of toluene / total pressure

=7.18 / 80.43 = 0.09

2) moles of sodium sulphate = 15.0 g /142.04 g = 0.106 mol

i= 3 ( as it will give three ions in solution)

deta T_f = 3 x1.86 x 0.106 = 0.59

freezing point = - 0.59 ⁰C