Two elements A and B form compounds having formula `AB_(2)` and `AB_(4)` . When dissolved in 20 g of benzene (`C_(6)H_(6)`) , 1g 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 atomic masses of A and B.

Explanation

Applying the formula, `M_(2)=(1000K_(f)w_(2))/(w_(1)xxDeltaT_(f))`
`M_(AB_(2))=(1000xx5.1xx1)/(20xx2.3)"110.87 g mol"^(1),M_(AB_(4))= (1000xx5.1xx1)/(20xx1.3)= "196.15 g mol"^(-1)`
Suppose atomic masses of A and B are 'a' and 'b' respectively. Then
Molar mass of `AB_(2)=a+2b=110.87" g mol"^(-1)" ...(i)"`
Molar mass of `AB_(4)=a+4b="196.15 g mol"^(-1)" ...(ii)"`
Eqn. (ii) - Eqn. (i) gives 2 b = 85.28 or b = 42.64
Substituting in eqn. (i), we get `a+2xx42.64=110.87 or a=25.59`
Thus, Atomic mass of A = 25.59 u, `" "` Atomic mass of B = 42.64 u