Question

The freezing point depression of a solution of 1.45 g sample of an unknown nonelectrolyte dissolved in 25.00 mL of benzene(density d=0.879 g/mL) is 1.28^oC.

Pure benzene has Kf value of 5.12°C/m.

What is the molecular weight of the compound? (10 points)

Answer 1

Freezing-point depression is the decrease of the freezing point of any paricular solvent on the addition of a misble non-volatile solute. It is expressed as:

∆T = imK_f [where, ∆T = Freezing-point depression (given), i = Van't Hoff factor (given); m = molality (to be determined); K_f = freezing constant for benzene (given)]

25.00 mL of benzene means = (25.00 mL x 0.879 g/mL) = 21.975 g = 0.021975 Kg of benzene

Considering the molecular weight of the unknown compound is M (g/mol)

Thus, converting the mol of the 1.45 g sample will be = 1.45 g / M (g/mol) = (1.45/M) mol

Thus, molality of the unknown compound will be = mol/amount of solvent used = [(1.45/M) mol] / 0.021975 Kg = 1.45/0.021975M = 65.984/M

Putting all the data in the ∆T = imK_f equation [assuming the Vant Hoff factor, i = 1 for the solute]

1.28 = 1 x (65.984/M) x 5.12

Thus M = (65.984 x 5.12)/1.28

M = molecular weight = 263.93 g/mol [answer]