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Complete reaction of $2.0 \mathrm{~g}$ of calcium (at. wt. = 40 ) with excess $\mathrm{HCL}$ produces $1.125 \mathrm{~L}$ of $\mathrm{H}_{2}$ gas. Complete reaction of the same quantity of another metal "M" with excess $\mathrm{HCL}$ produces $1.85 \mathrm{~L}$ of $\mathrm{H}_{2}$ gas under indentical conditions. The equivalent weight of "M" is closest to
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The correct answer is:
12
$1.125 \mathrm{~L}$ of $\mathrm{H}_{2}$ produced by $0.1$ equivalent of metal $1.85 \mathrm{~L}$ a of $\mathrm{H}_{2}$ will be produced by $=\frac{0.1 \times 1.85}{1.125}$ equivalents $\therefore$ No of gram equivalent of metal
$\begin{array}{l}
=\frac{2}{\text { Equivalent weight }}=\frac{2}{x} \\
\therefore \frac{0.1}{1.125} \times 1.85=\frac{2}{x} \\
x=12.16
\end{array}$
$\begin{array}{l}
=\frac{2}{\text { Equivalent weight }}=\frac{2}{x} \\
\therefore \frac{0.1}{1.125} \times 1.85=\frac{2}{x} \\
x=12.16
\end{array}$
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