A solution is made by dissolving 0.688 mol of nonelectrolyte solute in 807 g of benzene. Calculate the freezing point, T f , and boiling point, T b , of the solution.

Question

Kf=5.12 C/m

Kb=2.53 C/m

Answer 1

Freezing point (as well as boiling point, osmotic pressure, etc.) is considered a colligative property. It depends on the number of particles present in solution. The more particles, the lower the freezing point. In the present question, we have a nonelectrolyte, meaning it does not ionize or dissociate, so the number of particles is one.

For the freezing point depression, we use the equation ∆T = imK

∆T = change in freezing point

i = van't Hoff factor = # of particles = 1

m = molal concentration = moles of solute/kg solvent = 0.688 mol/0.807 kg = 0.853 m

K = freezing point constant = 5.12ºC/m

∆T = (1)(0.853 m)(5.12ºC/m) = 4.37ºC (note: this is the CHANGE in freezing point)

New freezing point of the the solution = normal freezing point of benzene - 4.37ºC. Look up the normal freezing point of benzene and subtract 4.37 degrees to get the answer.

Boiling point elevation:

∆T = imK

all variables are the same as above calculation, except we use Kb = 2.53ªC/m as the boiling point constant:

∆T = (1)(0.853 m)(2.53ºC/m) = 2.16ºC (note: this is the CHANGE in boiling point)

New boiling point of the solution = normal boiling point of benzene + 2.16ºC. Look up the normal boiling point of benzene and add 2.16 degrees to get the answer.