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`.

Explanation

We know that,
`m=(1000K_(f)xxw)/(WxxDeltaT)`
Molecular mass of `AB_(2)` (from given data) `= (1000xx5.1xx1)/(20xx2.3)`
`=110.86`
and Molecular mass of `AB_(4)` (from given data) `= (1000 xx 5.1 xx1)/(1.3 xx 20)`
`= 196.15`
Further, `AB_(4)=A+4B=196.15`....(i)
`AB_(2)=A+2B=110.86`....(ii)
Subtracting eq. (ii) from (i),
`A+85.29 = 110.86`
or `A= (110.86 - 85.29)=25.57`
Thus the atomic masses of A and B are 22.57 and 42.645 respectively.