What is the freezing point of an benzene solution that boils at 81.02 degrees C? Normal freezing point is 5.5 and the normal boiling point is 80.1 degrees C. Kfp = 5.12 K/m and Kbp = 2.53 K/m.

Question

What is the freezing point of an benzene solution that boils at 81.02 degrees C? Normal freezing point is 5.5 and the normal boiling point is 80.1 degrees C.

Kfp = 5.12 K/m and Kbp = 2.53 K/m.

Correct answer: 3.64

Answer 1

You are given the normal freezing point and normal boiling point of benzene. You are also given the increase in b.p. of the solution, so we can figure out how many molality of the solution. We can then calculate the new freezing point. First, we will have to make one assumption and that is that the solute that is raising the b.p. and lowering the f.p. is a non electrolyte and does not ionize. Thus, the van't Hoff factor will be 1.

For b.p. elevation = ∆T = imK where T = change in b.p. = 81.02 - 80.1 = 0.92; i = van't Hoff factor = 1 for a non electrolyte; m = molality = ? and K = b.p. constant = 2.53 K/m. Solving for m we have m = ∆T/(i)(K)

m = 0.92/(1)(2.53) = 0.36 m (ignoring the correct number of sig. figs.)

To find the new freezing point of this solution, we use ∆T = imK where K is now the freezing point constant.

∆T = (1)(0.36)(5.12) = 1.86 degrees

New freezing point = 5.5 - 1.86 = 3.64ºC

(NOTE: the K values are given as K/m but since the CHANGE is temperature is used, it makes no difference if you use K or C as the units, so a K of 2.53 K/m = 2.53ºC/m)