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

`Aw_(A)=25.58mu, Aw_(B)=42.64mu`

Explanation

`W_(2)(AB_(2))=1g,Mw_(2)(Ag_(2))=?`
First case
`W_(1)(C_(6)H_(6))=20g,Mw_(1)(C_(6)H_(6))=78.9g`
`DeltaT_(f)=2.3K,K_(f)=5.1`
`2.3=5.1xx(1xx1000)/(Mw_(2)(AB_(2))xx20) ……..(i)`
Second case
`Mw_(2)(AB_(2))=110.86g mol^(-1)`
`W_(2)(AB_(4))=1g,Mw_(2)(AB_(4))=?`
`W_(1)(C_(6)H_(6))=20g,DeltaT_(f)=1.3K`
`1.3=5.1xx(1xx1000)/(Mw_(2)(AB_(4))xx20) ......(ii)`
`Mw_(2)(AB_(4))=196.15g mol^(-1)`
Let the atomic weight of `A=x`
Let the atomic weight of `B=y`
`x+2y=110.86`
`x+4y=196.15`
Solve for `x` for `y`, we get
`y=42.64,x=25.58`
`:. ` Atomic weight of `A=25.58mu`
Atomic weight `B=42.64mu`