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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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

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$40.3$

$24.3$

$11.3$

$36.3$

answer is C.

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Detailed Solution

As the given sulphate is isomorphous with ${ZnSO}_{4} \cdot 7 H_{2} O$ its formula would be

${MSO}_{4} \cdot 7 H_{2} O \cdot m$ is the atomic weight of $M,$ Molecular weight of

${MSO}_{4} \cdot 7 H_{2} O = m + 32 + 64 + 126 = m + 222$

Hence $%$ of $M$ $= \frac{m}{m + 222} \times 100 = 9 .87$ (given) or

$\begin{array}{l} 100 m = 9 .87 m + 222 \times 9 .87 or 90 .13 m = 222 \times 9 .87 \\ or m = \frac{222 \times 9 .87}{90 .13} = 24 .3 \end{array}$

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