The depression of freezing point of a 2% aqueous solution of a substance A is equal to the depression of freezing point of a 4% aqueous solution of substance B. If the molecular weight of B is `122.5gmol^(-1)`, then the molecular weight of Ais `gmol^(-1)`.
(Assume that both A and B are non electrolytes)

Answer

`60`

Explanation

`DeltaT_(f)=K_(f)(W_(B))/(M_(B)W_(A))xx1000`
Let molecular weight of `A=x gmol^(-1)`
2% aqueous solution means 2 gof soute in 100g solution. Then `DeltaT_(f)("in"A)=(2K_(f))/(x xx98)xx1000`
4% aqueous solution means 4 g of solute in
100 g of solution
`W_(A)=100-4=96=96xx10^(-3)kg`
`DeltaT_(f)("in"B)=(4K_(r))/(122.5xx96)xx1000`
Since `DeltaT` is equal for two solutions,
`(2K_(f))/(x xx 98)xx1000=(4K_(f))/(122.5xx96)xx1000`
`x=(2xx122.5xx96)/(4xx98)`
`:.x=60gmol^(-1)`