Question

An aqueous solution freezes at -2.55 c whatvis its boiling point

Answer 1

Freezing point of an aqueous solutionis (-0.186)°C. Freezing point of an aqueous solution is (-0.186)°C. Elevation of boiling point of the samesolution is Kb = 0.512 °C, Kf = 1.86 °C, find the increase in boiling point.

Answer 2

Hey dear,

● Answer-
Tb' = 100.7019 °C

● Explaination -
# Given-
Tf' = -2.55 °C

# Solution-
Depression in freezing point -
∆Tf = 0 -(-2.55) = 2.55 °C

Boiling point elevation is calculated by -
∆Tb = ∆Tf × Kb / Kf
∆Tb = 2.55 × 0.512 / 1.86
∆Tb = 0.7019 °C

Thus, boiling point is -
Tb' = Tb + ∆Tb
Tb' = 100 + 0.7019
Tb' = 100.7019 °C

Therefore, boiling point of the solution is 100.7019 °C.

Hope it helps...