Question

An aqueous solution is 0.0222 m Determine the boiling point and freezing point of the solution. Kb=0.512°C/m K = 1.86°C/m 89.5 00:-0.015 00 0 100.01100:-0.04100 120 00:-0.0550C 11100-0.10100

Answer 1

Given, molality of aqueous solution (m) = 0.0222 m

Kb = 0.512° C/m

Kf = 1.86° C/m

Boiling point of water (T​​b°) = 100° C

Freezing point of water (T​​f°) = 0° C

Using the laws of elevation in boiling point and depression in freezing point, we get

$\Delta$Tb= Kb x m

Tb - Tb° = 0.512° C/m x 0.0222 m = 0.011° C

Tb = 100° C + 0.011° C = 100.011° C

$\Delta$Tf = Kf x m = 1.86° C/m x 0.0222 m = 0.041° C

Tf° - Tf = 0.041° C => Tf = 0° C - 0.041° C = -0.041° C

Answer: 100.011°C, -0.041° C (option second)

Let me know if you have any queries regarding this.