Two elements A and B form compounds having molecular formulae `AB_2 and AB_4` when dissolved in 20 g of `C_6H_6. 1g AB_2` lowers the freezing point by 2.3 K whereas 1.0 g of `AB_4` lowers it by 1.3 K. The molal depression constant for benzene is `5.1 K kg mol^(-1)`. The atomic masses of A and B are, respectively

Options

A. A = 26, B = 42.64
B. A = 31.72, B = 47.02
C. A=13.11, B=24.25
D. A = 19.17, B = 35.01

Answer

A

Explanation

Suppose atomic masses of A and B are x and y, respectively.
Molar mass of `AB_2 = x + 2y, 1g AB_2 = 1/(x + 2y)` mol
Molar mass of `AB_4 = x + 4y, 1 g AB_4 = 1/(x+4y)` mol
`m(AB_2)=(1//(x+2y))/20 xx1000 = 50/(x+2y)`
`DeltaT_f,=K_fxx m`
`2.3= 5.1xx50/(x+ 2y) implies x+2y=110.87` ...(i)
Similarly, for the `2^(nd)` case
`x+4y = 196.15` ...(ii)
Solving Eqs. (i) and (ii), we get x = 25.59, y = 42.64