A 0.2 molal aqueous solution of weak acid (HX) is 20% ionised. The freezing point of this solution is (Given, `K_(f) = 1.86^(@) C m^(-1)` for water)

Options

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

Answer

A

Explanation

Depression in freezing point, `Delta T_(f)=ixx K_(f)xx m`
Van't Hoff factor, `i=(1-alpha+n alpha)/(1)`, where n = no. of ions produced by complete dissociation of 1 mole of HX.
`HX hArr H^(+)+X^(-) rArr n = 2`
`therefore i=(1-0.2+2xx0.2)/(1)=1.2`
`therefore Delta T_(f)=1.2xx1.8xx0.2=0.432`
Hence freezing point of solution is `0-0.432 = -0.432`.