The degree of ionization of 0.2 m weak acid HX is 0.3. If the `K_(f)` for water = 1.85. What will be the freezing point of solution ?

Options

A. `-0.360^(@)C`
B. `-0.206^(@)C`
C. `+0.480^(@)C`
D. `-0.480^(@)C`

Answer

D

Explanation

`{:(,HX,hArr,H^(+),+,X^(-)),("Initial mole","1mole",,0,,0),("Final mole",(1-0.3),,0.3,,0.3):}`
Total mole `= (1-0.3)+0.3+0.3=1.3`
`i=(1.3)/(1)=1.3`
`Delta T_(f)=ixxmxxK_(f)`
`= 1.3xx1.85xx0.2`
`= 0.480^(@)C`
`therefore` Freezing point `= 0^(@)C-0.480^(@)C`
`= -0.480^(@)C`