16. An aqueous solution has a normal boiling point of 103.5°C. What is the freezing point of this solution? for water Kb=0.51

Question

For water Kb=0.51 degrees C/m Kf=1.86 degress C/m

Answer 1

2 = (1) (0.51) (m)

m = 3.92

Δt = (1) (1.86) (3.92)

Δt = 7.29 C

The freezing point is -7.3C.

Try it with a van 't Hoff factor of 2, you'll get the same result. Since it is the same solution in both cases, the van 't Hoff factor is the same in each equation.

Answer 2

Find the molality of the solution:

∆T = imK_b

∆T = change in boiling point = 3.5ºC

i = van't Hoff Factor = 1 assuming a non-electrolyte as the solute (it isn't stated)

m = molality = moles solute/kg solvent

K_b = boiling point constant = 0.51

m = (3.5)/(1)(0.51) = 6.86 m

 

Now substitute this value into the freezing point expression and solve for ∆T:

∆T = imK_f

∆T = (1)(6.86)(1.86)

∆T = 12.8ºC

So, the freezing point of the solution will be -12.8ºC