A `0.2` molal aqueous solution of a weak acid `HX` is `20%` ionized. The freezing point of the solution is `(k_(f) = 1.86 K kg "mole"^(-1)` for water):

Options

A. `-0.45^(@)C`
B. `-0.90^(@)C`
C. `-0.31. ^(@)C`
D. ` 0.53 .^(@)C`

Answer

A

Explanation

ACQ 20 % ionized
`alpha = 20/100 = 0.2 `
Vant hoff factor I = `1 + alpha rArr I = 1 +0.2 = 1.2 `
`DeltaT_(f) = I xx K_(f) xx m `
`DeltaT_(f) = 0.2 xx 1.2 xx 1.86 = 0.45`
Therefore freezing point = `- 0.45 . ^(@)C` .