Two elements `A` and `B` form compounds having molecular formula `AB_(2)` and `AB_(4)`. When dissolved in `20 g` of benzene, `1 g`of `AB_(2)` lowers the freezing point by `2.3 K`, whereas `1.0 g` of `AB_(4)` lowers it by `1.3 K`. The molar depression constant for benzene is `5.1 K kg mol^(-1)`. Calculate the atomic mass of `A` and `B`.

Answer

`25.57g "mole"^(-1),42.645g "mol"^(-1)`.

Explanation

DeltaT_(f)=K_(f)xx(W_(B))/(M_(B))xx(1000)/(W_(A))`
`2.3K=5.1xx(1)/(M_(B))xx(1000)/(20) rArr M_(B)=(5.1xx50)/(2.3)=(255)/(2.3)=110.86`
`1.3K=5.1xx(1)/(M_(B))xx(1000)/(20) rArr M_(B)=(5.1xx50)/(1.3)=(255)/(1.3)xx196.15`
`A+2B=110.86`
`A+4B=169.15`
`-2B=-85.29`
`B=42.645g "mol"^(-1)`
`A+2B=110.86`
`rArr A+2xx42.645=110.86`
`A=110.86-85.29=25.57g "mol"^(-1)`