Question

A solution is made by dissolving 0.592 mol of nonelectrolyte solute in 767 g of benzene. Calculate the freezing point, Te, and boiling point, Tb, of the solution. Constants can be found in the table of colligative constants. T =

Answer 1

first calculate molality of the solution

molality = n / mass of solvent in g

molality = 0.592 / 767

molality = 0.00077

now for Tf

$\Delta$Tf = Kf x m

$\Delta$Tf = 5.12 x 0.00077

$\Delta$Tf = 0.0039

Tf = 5.49 - 0.0039

Tf = 5.486 ⁰C

for Tb

$\Delta$Tb = Kb x m

$\Delta$Tb = 2.53 x 0.00077

$\Delta$Tb = 0.0019

Tb = 80.1 + 0.0019

Tb = 80.102⁰C