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Upon fully dissolving $2.0 \mathrm{~g}$ of a metal in sulfuric acid, $6.8 \mathrm{~g}$ of the metal sulfale is formed. The equivalent weight of the metal is $-$
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The correct answer is:
$20.0 \mathrm{~g}$
Equivalents of metal = Equivalents of metal sulphate
$\frac{\text { wt.of metal }}{\text { Eq.wt.of metal }} \frac{\text { wt.of metal sulphate }}{\text { Eq. wt.metal sulphate }}$
$\begin{array}{l}
\frac{2}{x}=\frac{6.8}{x+48} \\
6.8 x=2 x+96 \\
4.8 x=96 \\
x=\frac{96}{4.8}=20
\end{array}$
$\frac{\text { wt.of metal }}{\text { Eq.wt.of metal }} \frac{\text { wt.of metal sulphate }}{\text { Eq. wt.metal sulphate }}$
$\begin{array}{l}
\frac{2}{x}=\frac{6.8}{x+48} \\
6.8 x=2 x+96 \\
4.8 x=96 \\
x=\frac{96}{4.8}=20
\end{array}$
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