Question

An aqueous solution freezes at -2.55°C . What is its boiling point?
[Kb for water= 0.52
Kf for water=1.86]

Answer 1

Answer:

As the solution freezes at -0.186, Its Depression in Freezing point(ΔTf) is 0.186.

(Kf = 1.86 is called Cryoscopic constant)

→ ΔTf = Kf x m

→ 0.186 = 1.86 x m

→ m = 0.1

So, molality of the aqueous solution = 0.1

Elevation in boiling point,

(Kb =0.52 is called Ebullioscopic constant)

→ ΔTb = Kb x m

→ ΔTb = 0.52 x 0.1

→ ΔTb = 0.052

Thus the Elevation in Boiling point is 0.052