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.

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 .

Two elements A and B form compounds having molecular formula AB_(2) and AB_(4) . When dissolved in 20g of benzene, 1g of AB_(2) lowers the fpt by 2.3K whereas 1g of AB_(4) lowers it by 1.3K (K_(f) for C_(6)H_(6) = 5.1 K m^(-1)) At. Mass of A and B are

The elements X and Y form compound having molecular formula XY_(2) and XY_(4) (both are non - electrolysis), when dissolved in 20 g benzene, 1 g XY_(2) lowers the freezing point by 2.3^(@)c whereas 1 g of XY_(4) lowers the freezing point by 1.3^(C) . Molal depression constant for benzene is 5.1. Thus atomic masses of X and Y respectively are

1.0 g of non-electrolyte solute dissolved in 50.0 g of benzene lowered the freezing point of benzene by 0.40 K . The freezing point depression constant of benzene is 5.12 kg mol^(-1) . Find the molecular mass of the solute.