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

`underset("m(1-0.20)")(HX) hArr underset("m(0.20)")(H^(+)) + underset("m(0.20)")(X^(-))`
Molal conc. In solution.
` = m (1-0.20) + 0.20 m + 0.20 m`
` = 1.20 m`
and ` " " m = 0.2`
`Delta T_(f) = K_(f) xx m`
` = 1.86 xx 1.20 (0.2)`
` = 0.446 = 0.45`
`therefore ` Freezing point of solution ` = - 0.45^(@)C`