A 0.1 molal aqueous solution of a weak acid is 30% ionised. If `K_f` for water is `1.86^@C//m`, the freezing point of the solution will be

Options

A. `- 0.18^@C`
B. `- 0.54^@C`
C. `- 0.36^@C`
D. `- 0.24^@C`

Answer

D

Explanation

We know that `DeltaT_f= i xx K_f xx m`
Here i is van.t Hoff.s factor. Here a is degree of dissociation i.e., 30% i.e., 30/100 = 0.3
`:. i= 1 + alpha = 1 + 0.3 = 1.3`,
`DeltaT_f = ixx K_f xx m = 1.3 xx 1.86 xx 0.1 = 0.24`
`:. " Freezing point " = -0.24^@C`