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  4. The sulphate of a metal M contains 9.87% of M. This sulphate is isomorphous with ZnSO4.7H2O. The atomic weight of M is

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Q. The sulphate of a metal $M$ contains 9.87% of $M$. This sulphate is isomorphous with $ZnSO_4.7H_2O$. The atomic weight of $M$ is

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Solution:

As the given sulphate is isomorphous with $ZnSO_4.7H_2O$ its formula Would be $MSO_4.7H_2O.m$ is the atomic weight of $M$ molecular weight of $MSO_4.7H_2O$
$= m + 32 + 64 + 126 = m + 222$
Hence % of $M=\frac{m}{m+222}\times100=9.87$(given) or
$100m = 9.87m + 222 \times 9.87 or 90.13m = 222 \times9.87$
or $m=\frac{222\times9.87}{90.13}=24.3$