Question

A 0.1 molal aqueous solution of a weak acid is 30% ionized.
If Kf for water is 1.86°C/m, the freezing point of the solution
will be :
(a) – 0.18°C (b) – 0.54°C (c) – 0.36°C (d) – 0.24°C

Answer 1

The freezing point of the solution is ΔTf = - 0.24

Explanation:

We know that

ΔTf = i x  Kf  x m

Here i is van't Hoff's factor. i for weak acid is 1 + α.

Here α is degree of dissociation i.e. 30/100 = 0.3

i = 1 + α = 1 + 0.3 = 1.3

ΔTf = i x  Kf  x m

ΔTf = 1.3 x 1.86 x 0.1

ΔTf = 0.24

Freezing point = - 0.24

Thus the freezing point of the solution is ΔTf = - 0.24

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Calculate the freezing point of a solution when 3 gram of calcium chloride m equals to 100 gram per mole was dissolved in 100 gram of water assuming CaCl2 undergoes complete ionization kf for water equals to 1.86 k kg per mole  ?