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Q)

A metallic oxide contains 60% of the metal.The equivalent weight of the metal is

$\begin{array}{1 1}(a)\;12& (b)\;24\\(c)\;40&(d)\;48\end{array} $

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A)
Eq.wt=$\large\frac{\text{Wt of metal}}{\text{Wt of oxygen combined}}$$\times 8$
$\Rightarrow \large\frac{60}{40}$$\times 8=12$
Hence (a) is the correct option.
But how the formula can be established???
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