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)), 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 atomic masses of A and B.

Two elements A and B form compounds having formula AB_(2) and AB_(4).

1.	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)), 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 atomic masses of A and B.
Answer» Solution :We know that, `M_(2)=(1000xx w_(2)xx K_(f))/(Delta T_(f)xx w_(1))` Then, `M_(AB_(2))=(1000xx1xx5.1)/(2.3xx20)=110.87g mol^(-1)` `M_(AB_(4))=(1000xx1xx5.1)/(1.3xx20)=196.15 g mol^(-1)` Now, we have molar masses of `AB_(2)` as `110.87 g mol^(-1)`and `196.15 g mol^(-1)` RESPECTIVELY. Let the atomic masses of A and B x and y respectively. Now, we can write : `x + 2y=110.87 ""`.....(i) `x + 4y=196.15 ""`....(ii) SUBTRACTING EQUATION (i) from (ii), we have `2y=85.28` `therefore y=42.64` Putting the value of y in equation (1), we have `x + 2xx42.64 = 110.87` `therefore x = 85.28` Hence, the atomic masses of A and B are 25.59 u and 42.64 u respectively.