Question

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 K whereas 1.0 g of AB₄ lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg mol⁻¹. Calculate the atomic masses of A and B.

Answer 1

We know that,

M₂ = `(1000 xx w_2 xx K_f)/(Delta T_f xx w_1)`

Then `M_(AB_2) = (1000 xx 1 xx 5.1)/(2.3 xx 20)`

= 110.87 g mol⁻¹

`M_(AB_4) = (1000 xx 1 xx 5.1)/(1.3 xx 20)`

= 196.15 g mol⁻¹

Now, the molar masses of AB₂ and AB₄ are 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,

2y = 85.28

⇒ y = 42.64

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

x + 2 × 42.64 = 110.87

x + 85.28 = 110.87

x = 110.87 − 85.28

⇒ x = 25.59

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