Depression in freezing point of 0.1 molal solution of HF is `-0.201^(@)C`. Calculate percentage degree of dissociation of HF. `(K_(f)=1.86 K kg mol^(-1))`.

Answer

`26.5%`

Explanation

The observed molecular mass can be calculated from the reaction.
`M_(B) = (K_(f) xx 1000 xx w_(B))/(w_(A) xx Delta T_(f))`
0.01 m aquous solution means 0.01 mole or `0.01 xx 142 `( mol wt of `Na_(2) SO_(4)` = 142) g of sodium sulphate are dissolved in 1000g of water.
`w_(B) = 1.42 g , w_(A) = 1000g , Delta T_(f) = 0.0284`
`M_(B) `(Observed) ` = (1.86 xx 1000 xx 1.42)/(1000 xx 0.0284) = 93`
Vant. Hoff factor , `i= ("Normal mol.mass")/("Observed mol.mass")`
`i= (142)/(93) = 1.53`
Degree of dissociation can be calculated as :
`Na_(2)SO_(4) hArr2N^(+) + SO_(4)^(2-)`
Initial moles `" "1" "0" "0`
After dissociation `1-alpha" "2 alpha" "alpha`
Total number of moles after dissociation `= 1 - alpha + 2alpha + alpha = 1+ 2alpha`
` i = ("Moles after dissociation")/("Normal moles")`
` i = (1+2alpha)/(1) = 1.53`
`therefore " " (1+2alpha)/(1) = 1.53`
or `" " 2alpha = 1.53`
or `" " 2 alpha = 1.53 -1 =0.53`
`alpha = 0.53//2 = 0.256`
`therefore` Degree of dissociation = 26.5% .