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

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(C) Since the given sulphate is isomorphous with $ZnSO_4 \cdot 7H_2O$,its formula is $MSO_4 \cdot 7H_2O$.Let $m$ be the atomic weight of $M$.The molec

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$24$

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(C) Since the given sulphate is isomorphous with $ZnSO_4 \cdot 7H_2O$,its formula is $MSO_4 \cdot 7H_2O$.Let $m$ be the atomic weight of $M$.The molecular weight of $MSO_4 \cdot 7H_2O = m + 32 + (4 	imes 16) + 7 	imes (2 	imes 1 + 16) = m + 32 + 64 + 126 = m + 222$.The percentage of $M$ is given by $\frac{m}{m + 222} 	imes 100 = 9.87$.$100m = 9.87(m + 222)$.$100m = 9.87m + 2191.14$.$90.13m = 2191.14$.$m = \frac{2191.14}{90.13} \approx 24.3$.

The sulphate of a metal $M$ contains $9.87 \%$ of $M$. This sulphate is isomorphous with $ZnSO_4 \cdot 7H_2O$. The atomic weight of $M$ is: (in $.3$)

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