Two elements A and B form compounds having molecular formulae AB2 and AB4, when dissolved in 20 g of C6H6. 1.0 g of AB2 lowers the freezing point by 2.3 K whereas 1.0 g of AB4 lowers it by 1.3 K. The molal depression constant for benzene is 5.1 K kg mol-l. The atomic masses of A and B are, respectively

Question

Two elements A and B form compounds having molecular formulae AB2 and AB4, when dissolved in 20 g of C6H6. 1.0 g of AB2 lowers the freezing point by 2.3 K whereas 1.0 g of AB4 lowers it by 1.3 K. The molal depression constant for benzene is 5.1 K kg mol-l. The atomic masses of A and B are, respectively (A) 26, 42.64 (B) 31.72, 47.02 (C) 13.11, 24.25 (D) 19.17, 35.01

Tardigrade Admin · Accepted Answer

Suppose atomic masses of A and B are x and y respectively.
Molar mass of

A B_{2} = x + 2 y

(x + 2 y) g

of

A B_{2} = 1 m o l

1 g

of

A B_{2} = \frac{1}{x + 2 y} m o l

Molar mass of

A B_{4} = x + 4 y

(x + 4 y) g

of

A B_{4} = 1 m o l

1 g

of

A B_{4} = \frac{1}{x + 4 y} m o l

m_{(x + 4 y)} = \frac{1/ ( x + 2 y )}{20} \times 1000 = \frac{50}{x + 2 y}

Δ T_{f} = K_{f} \times m

2.3 = 5.1 \times \frac{50}{x + 2 y}

or

x + 2 y = 110.87 \dots (i)

Similarly, for the second case

m_{(A B_{4}) =} = \frac{1/ ( x + 4 y )}{20} \times 1000 = \frac{50}{x + 4 y}

Δ T_{f} = K_{f} \times m

1.3 = 5.1 \times \frac{50}{x + 4 y}

x + 4 y = 196.15 \dots (ii)

Solving eqs

(i)

and

(ii)

, we get

x = 25.59; y = 42.64