Two elements A and B form compounds having formula

Solution:

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.