Question

Compare the equations for calculating freezing point depression and boiling point elevation. Illustrate the difference by calculating the ΔT for the freezing point and boiling point of 2.0 molality NaCl.

Answer 1

Compare the equations for calculating freezing point depression and boiling point elevation. Illustrate the difference by calculating the ΔT for the freezing point and boiling point of 2.0 molality NaCl.

For freezing point, we will expect a depression

and for boiling point, we shoudl expect a higher point, i.e elevation

so

T mix freezing = Tf - dTf

T mix boiling = Tb + dTb

so

dTf = -Kf*m*i; where Kf = depression point constant, m = molality, i = ions in solution

dTb = Kb*m*i = where Kb = elevation point constant, m = molality, i = ions in solution

then

For NaCl

molal = 2

i = 2 ions, so

For water Kf = 1.86 and Kb = 0.512

dTb = Kb*m*i = 0.512*2*2 = 2.05 °C increases,

Tb mix = 100+2.05 = 102.5°C

freezing:

dTf = -Kf*m*i

dTf = -1.86*2*2 = -7.44

Tf mix = 0 -7.44 = - 7.44° C