`1 g` of monobasic acid in `100 g` of water lowers the freezing point by `0.168^(@)`. If `0.2 g` of same acid requires `15.1 mL mol^(-1)` of `N//10` alkali for complete neutralization, calculate the degree of dissociation of acid. `K'_(f)` for `H_(2)O` is `1.86 K mol^(-1) kg`.

Options

A. `9.8%`
B. `19.6%`
C. `4.9%`
D. `1.68%`

Answer

B

Explanation

`M_(2)" (mol mass of acid)"=(1000xxK_(f)xxw_(2))/(w_(1)xxDeltaT_(f))`
`=(1000xx1.86xx1)/(100xx0.168)=110.71`
i.e., Observed molecular mass = 110.71
15.1 mL of `(N)/(10)` alkali neutralise acid = 0.2 g
`=(0.2)/(15.1)xx1000xx10=132.45g`
`therefore" Eq. wt. of the acid = 132. 45"`
Mol. mass of the acid = 132.45 (as it is monobasic)
This is calculated (normal) mol. mass
`i=("Normal mol. mass")/("Observed mol. mass")=(132.45)/(110.71)=1.196`
`{:(,HA ,hArr,H^(+),+,A^(-)),("Inital",1,,0,,0),("After disso. ",1-alpha,,alpha,,alpha),(,,,,,"Total "=1+alpha):}`
`therefore i=1+alpha or alpha=i-1=0.196=19.6%`