Question

the freezing point of an aqueous solution is - 0.186°C. what is the elevation of boiling point of this solution if Kf=1.86 and Kb= 0.512?

Answer 1

Answer:

Explanation:

Depression in freezing point (ΔTf) = Kf x m

0.186 = 1.86 x m

m = 0.1

Now as we have got the molality, m= 0.1

Elevation in boiling point (ΔTb) = Kb x m

→ ΔTb = 0.512 x 0.1

→ ΔTb = 0.0512

Therefore, the elevation in Boiling point is 0.0512.

Hope it helps!!

Answer 2

Answer:

$0.0512$

Explanation:

It is the correct answer.