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 $B$.

(N/A) We know that the formula for molar mass is $M_{2} = \frac{1000 \times w_{2} \times K_{f}}{\Delta T_{f} \times w_{1}}$.
For $AB_{2}$: $M_{AB_{2}} = \frac{1000 \times 1 \times 5.1}{2.3 \times 20} = 110.87 \ g \ mol^{-1}$.
For $AB_{4}$: $M_{AB_{4}} = \frac{1000 \times 1 \times 5.1}{1.3 \times 20} = 196.15 \ g \ mol^{-1}$.
Let the atomic masses of $A$ and $B$ be $x$ and $y$ respectively.
Then,$x + 2y = 110.87$ $(i)$ and $x + 4y = 196.15$ $(ii)$.
Subtracting $(i)$ from $(ii)$,we get $2y = 85.28$,so $y = 42.64 \ u$.
Substituting $y$ in $(i)$,$x + 2(42.64) = 110.87$,so $x = 25.59 \ u$.
The atomic masses of $A$ and $B$ are $25.59 \ u$ and $42.64 \ u$ respectively.