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

Let us first calculate molar mass of `AB_(2)` and `AB_(4)` .
For `AB_(2)` compound .
`M_(B) = (k_(f) xx w_(B) xx 1000)/(w_(A) xx Delta T_(f))`
`Delta T_(f) = 2.3 K , w_(B) = 1.0 g, w_(A)= 20.0g`
`K_(f) = 5.1 K kg mol^(-1)`
`M_(AB_(2)) = (5.1 xx 1.0xx 1000)/(20.2 xx 2.3) = 110.87`
`therefore " " M_(AB_(4)) = 196.15`
Let a is the atomic mass of A and B is the atomic mass of B, then
`M_(AB_(2)) = a + 2b + = 110.87 " ".....(i)`
`M_(AB_(4)) = a+ 4b = 196.15 " "........(ii)`
Subtracting eqn. (ii) from eqn.(i)
` - 2b = - 85.28`
`therefore " " b = 42.64`
Substituting value of b in eqn (i)
`a+2 xx 42.64 = 110.87`
`a = 110.87 - 85.28 = 25.59`
Atomic mass of A = 25.25
Atomic mass of D = 42.64.