The sulphate of a metal M contains 9.87 % of | vedclass

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(C) Since the given sulphate is isomorphous with $ZnSO_4 \cdot 7 H_2 O$,its formula is $MSO_4 \cdot 7 H_2 O$.Let $m$ be the atomic weight of $M$.The m

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$24$

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(C) Since the given sulphate is isomorphous with $ZnSO_4 \cdot 7 H_2 O$,its formula is $MSO_4 \cdot 7 H_2 O$.Let $m$ be the atomic weight of $M$.The molecular weight of $MSO_4 \cdot 7 H_2 O = m + 32 + (4 	imes 16) + 7 	imes (2 + 16) = m + 32 + 64 + 126 = m + 222$.Given that the percentage of $M$ is $9.87 \%$,we have:$\frac{m}{m + 222} 	imes 100 = 9.87$$100m = 9.87m + 222 	imes 9.87$$100m - 9.87m = 2191.14$$90.13m = 2191.14$$m = \frac{2191.14}{90.13} \approx 24.3$.Thus,the atomic weight of $M$ is $24.3$.

The sulphate of a metal $M$ contains $9.87 \%$ of $M$. This sulphate is isomorphous with $ZnSO_4 \cdot 7 H_2 O$. The atomic weight of $M$ is (in $.3$)

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