A metal M sulfate contains 9.87 % , M. If it | vedclass

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(C) Since $MSO_4 \cdot 7H_2O$ is isomorphous with $ZnSO_4 \cdot 7H_2O$,the molar mass of $MSO_4 \cdot 7H_2O$ is $M + 32 + (4 	imes 16) + 7 	imes (2

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$24$

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(C) Since $MSO_4 \cdot 7H_2O$ is isomorphous with $ZnSO_4 \cdot 7H_2O$,the molar mass of $MSO_4 \cdot 7H_2O$ is $M + 32 + (4 	imes 16) + 7 	imes (2 	imes 1 + 16) = M + 32 + 64 + 126 = M + 222 \, g/mol$.The percentage of metal $M$ is given by: $\frac{M}{M + 222} 	imes 100 = 9.87$.$100M = 9.87M + 222 	imes 9.87$.$100M - 9.87M = 2191.14$.$90.13M = 2191.14$.$M = \frac{2191.14}{90.13} \approx 24.3 \, g/mol$.

$A$ metal $M$ sulfate contains $9.87 \% \, M$. If it is isomorphous with $ZnSO_4 \cdot 7H_2O$,what is the atomic mass of $M$ (in $.3$)?

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