#### Question

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 Kwhereas 1.0 g of AB_{4} lowers it by 1.3 K. The molar depression constant for benzene is 5.1 Kkg mol^{−1}. Calculate atomic masses of A and B.

#### Solution

We know that,

`M_2 = (1000xxw_2xxk_f)/(triangleT_fxxw_1)`

Then `M_(AB_2) = (1000xx1xx5.1)/(2.3 xx 20)`

= 110.87 gmol^{-1}

`M_(AB_4) = (1000xx 1xx5.1)/(1.3xx20)`

= 196.15 g mol^{−1}

Now, we have the molar masses of AB_{2} and AB_{4} as 110.87 g mol^{−1} and 196.15 g mol^{−1} respectively.

Let the atomic masses of A and B be *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

2*y* = 85.28

⇒ *y* = 42.64

Putting the value of ‘*y*’ in equation (1), we have

*x* + 2 × 42.64 = 110.87

⇒ *x* = 25.59

Hence, the atomic masses of A and B are 25.59 u and 42.64 u respectively.