Question

An aqueous solution has a normal boiling point of 103 c. What is the freezing point of his solution? For Water Kb= 0.51 C/m and Kf= 1.86 C/,


I want the answer with datels please!!

Answer 1

elevation in boiling point , ?Tb = Kb X m
where ?Tb 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 ?Tb = 103- 100 = 3 degree centigrade...
Kb for water is 0.512 degree C/m

so putting the values and finding molality...
3 = 0.512 X m
m = 3/0.512 = 5.8594..

now depression in freezing point ?Tf = Kf X m
where ?Tf 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 5.8594
so ?Tf = 1.86 X 5.8594 = 10.8985 degrees c

now as normal water freezes at 0 degree centigrade but this solution has lowered the freezing point of solution by 10.8985 degree centigrade ....so freezing point of solution is 0 - 10.8985 = -10.8985 degree centigrade

Answer 2

elevation in boiling point = K_b*molality of the solution

now, boiling point of pure water = 100 degree C

and   boiling point of solution = 103 degree C

therefore, elevation in boiling point = 3 degree C

thus, 3 = 0.51*molality of the solution

or, molality of the solution = 5.882 m.............(1)

now, depression in freezing point = K_f*molality of the solution = 1.86*5.882 = 10.941

therefore, freezing point of solution = -10.941 degree C

************  freezing point of pure water = 0 degree C

Answer 3