Question

The boiling point of an aqueous solution is 102.34 °C. What is the freezing point?

Answer 1

Answer:
elevation in boiling point , ΔTb = Kb X m
where ΔTb is the elevation in boiling point...
Kb is molal elevation constant ...
and m is molality...

normal water boils at 100 degree centigrade but this aqueous solution boils at 105.9 degree centigrade ....so elevation in boiling point ΔTb = 102.34- 100 = 2.34 degree centigrade...
Kb for water is 0.512 degree C/m

so putting the values and finding molality...
2.34 = 0.512 X m
m = 2.34/0.512 = 4.57

now depression in freezing point ΔTf = Kf X m
where ΔTf is the depression in freezing point ..
Kf is the molal depression constant...
and m is molality..

Kf for water is 1.86 degree C/m
and moality as calculated above is 4.57
so ΔTf = 1.86 X 4.57 = 8.5
now as normal water freezes at 0 degree centigrade but this solution has lowered the freezing point of solution by 8.5 degree centigrade ....so freezing point of solution is 0 – 8.5 = -8.5 degree centigrade