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1g of a monobasic acid dissolved in 200g...

`1g` of a monobasic acid dissolved in `200g` of water lowers the freezing point by `0.186^(@)C`. On the other hand when `1g` of the same acid is dissolved in water so as to make the solution `200mL`, this solution requires `125mL` of `0.1 N NaOH` for complete neutralization. Calculate `%` dissociation of acid ? `(K_(f)=1.86(K-kg)/("mol"))`

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Verified by Experts

The correct Answer is:
60
Let molar mass of acid be m. then
0.186=`ixx1/mxx1000/200xx1.86`
Also, `1/m=125xx0.1xx10^(-3) so m=10000/125=400/5=80`
so `i=80/50=8/5=1.6=1+ alpha implies alpha=60%`
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