Complete reaction of 2.0 , g of calcium (at. | vedclass

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(D) . Let the equivalent weight of metal $M = x$. According to the law of equivalence,the number of equivalents of metal reacted is equal to the numbe

Vedclass · Accepted Answer

$12$

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(D) . Let the equivalent weight of metal $M = x$. According to the law of equivalence,the number of equivalents of metal reacted is equal to the number of equivalents of $H_2$ gas produced. $(eq)_{Ca} = (eq)_{H_2} 	ext{ (from Ca)} = \frac{2.0}{20} = 0.1$ $(eq)_{M} = (eq)_{H_2} 	ext{ (from M)} = \frac{2.0}{x}$ Since the volume of gas produced is directly proportional to the number of equivalents under identical conditions: $\frac{(eq)_{Ca}}{(eq)_{M}} = \frac{V_{H_2} 	ext{ (from Ca)}}{V_{H_2} 	ext{ (from M)}}$ $\frac{0.1}{2.0/x} = \frac{1.125}{1.85}$ $\frac{0.1x}{2.0} = \frac{1.125}{1.85}$ $x = \frac{1.125 	imes 2.0}{1.85 	imes 0.1} = \frac{2.25}{0.185} \approx 12.16$ Thus,the equivalent weight of $M$ is approximately $12$.

Complete reaction of $2.0 \, g$ of calcium (at. wt. $= 40$) with excess $HCl$ produces $1.125 \, L$ of $H_2$ gas. Complete reaction of the same quantity of another metal $M$ with excess $HCl$ produces $1.85 \, L$ of $H_2$ gas under identical conditions. The equivalent weight of $M$ is closest to

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