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

Given values are:
`(W_(AB))_2= 1.0 g , W_("benzene")=20 g`
`(W_(AB))_4=1.0 g`
`DeltaT_(f(AB_2)=2.3 K`
`DeltaT_(f(AB_4)=1.3 K`
Let us calculate the molar masses of `AB_(2)` and `AB_(4)`.
For `AB_(2)` Compound
`Mw_("solute")=(K_(f)xxW_("solute")xx1000)/(W_("solvent")xxDeltaT_(f)`
`Mw_(AB)_2 =(5.1xx1.0xx1000)/(20.0xx2.3)=110.87 g mol^(-1)`
`:. Mw(AB)_2 = 110.87 g mol^(-1)`
For `AB_(4)` compound
`Mw_(AB)_4 =(5.1xx1.0xx1000)/(20.0xx1.3)=196.15 g mol^(-1)`
Let `alpha` is the atomic mass of `A` and `beta` is the atomic mass of `B`, then ,
`Mw_(AB)_2 = alpha +2beta =110.87` ... (i)
`Mw_(AB)_4 = alpha +4beta =196.15` ... (ii)
Subtracting Eq. (ii) from Eq. (i), we get
`-2beta=-85.28`
`:. beta=42.64`
Substituting the value of `beta` in Eq.(i),
`alpha+2xx42.64=110.87`
`alpha=110.87-85.28=25.59`
Atomic mass of `A = 25.59`
Atomic mass of `B = 42.64`