A 0.2 molar aqueous solution of a weak acid (HX) is `20%` ionised . The freezing point of the solution is:
`(Given: K_(f)=1.86^(@) C kg" "mol^(-1)" for water")`

Options

A. `- 0.45`
B. `-0.90`
C. `-0.31`
D. `- 0.53`

Answer

A

Explanation

`Delta T_(f) = iK_(f)m` and `I = ( 1- alpha +nalpha ) /( 1 ) `
When n = number of ions formed on complete dissociation of 1 mole of HX.
`HX hArr H^(+) + X^(-) `i.e., n =2
`:. i = ( 1-0.2 + 2 xx 0.2)/( 1) = 1.2 `
or `Delta T_(f) 1.2 xx1.8 xx0.2 = 0.432`
Thus the freezing point of soluion `= 0-0.432= -0.432^(@)C `