Question

The sulphate of a metal M contains 9.87% of M. This sulphate is isomorphous with $ZnSO_{4}.7H_{2}O$. The atomic weight of M is

• A. 40.3
• B. 36.3
• C. 24.3
• D. 11.3

Answer 1

Answer: (C)

24.3

Answer 2

As the given sulphate is isomorphous with $ZnSO_{4}.7H_{2}O$ its formula would be $MSO_{4}.7H_{2}O$.m is the atomic weight of M, molecular weight of $MSO_{4}.7H_{2}O$ $= m + 32 + 64 + 126$ $= m + 222$

Hence % of M$= \frac{m}{m + 222} \times 100 = 9.87$ (given) or

$100m = 9.87m + 222 \times 9.87$or $90.13m = 222 \times 9.87$

or $m = \frac{222 \times 9.87}{90.13} = 24.3$.