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 Kg mol-1. Calculate atomic masses of A and B.
Given,
M2 = \(\frac{1000 \times w_2 \times k_f}{\Delta T_f \times w_1}\)
Then,
MAB2 = \(\frac{1000 \times 1 \times 5.1}{2.3 \times 20}\)
= 110.87 g mol-1
MAB4 = \(\frac{1000 \times 1 \times 5.1}{1.3 \times 20}\)
= 196.15 g mol-1
Now, 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 + 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 (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.