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.

Explanation

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.