Two elements A and B form compounds having formula AB2 and AB4. When dissolved in 20 g of benzene (C6H6), 1 g of AB2 lowers the freezing point by 2.3 Kwhereas 1.0 g of AB4 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.
We know that,
$M_{2}=\frac{1000 \times w_{2} \times k_{f}}{\Delta T_{f} \times w_{1}}$
Then, $M_{\mathrm{AB}_{2}}=\frac{1000 \times 1 \times 5.1}{2.3 \times 20}$
= 110.87 g mol−1
$M_{\mathrm{AB}_{4}}=\frac{1000 \times 1 \times 5.1}{1.3 \times 20}$
= 196.15 g mol−1
Now, we have the molar masses of AB2 and AB4 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:
$x+2 y=110.87$ ....(i)
$x+4 y=196.15$ ....(ii)
Subtracting equation (i) from (ii), we have
2y = 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.