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Question

The addition of nonvolatile solute results in a lowering of

the vapor pressure of the solvent.If 7.02 g of a nonionic

solute is dissolved in 82.30 g of benzene, C6H6, the freezing

point is 2.70oC. Find the molar mass of this solute. The

freezing point of benzene is 5.51oC and Kf is 5.12oC/m.

Answer 1

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The change in freezing point is given by ∆T = imK

∆T = change in freezing point = 5.51º - 2.70º = 2.81º

i = van't Hoff factor = 1 for nonionic substance

m = molality of solute = moles solute / kg benzene = ?

K = freezing constant for benzene = 5.12ºC / m

2.81 = (1)(m)(5.12)

m = 0.5488 moles solute / kg benzene

kg benzene = 82.30 g x 1 kg / 1000 g = 0.08230 kg

moles solute = 0.5488 moles / kg x 0.08230 kg = 0.0452 moles solute

Molar mass solute = g / moles = 7.02 g / 0.0452 mols = 155 g / mole