Ncert Exercises & Solutions

2.21 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 atomic masses of A and B.

We know that,

M_{2} = \frac{1000 \times w_{2} \times k_{f}}{∆ T_{f} \times w_{1}}

Then,

M_{{AB}_{2}} = \frac{1000 \times 1 \times 5.1}{2.3 \times 20} = 110.87 g {mol}^{- 1}

M_{{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.

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