0.1 g of metal combines with 46.6 mL of oxy | vedclass

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(A) At $STP$,$1 \ mol$ of $O_2$ occupies $22400 \ mL$ and provides $4 \ equivalents$ of oxygen (since $O_2 + 4e^- \rightarrow 2O^{2-}$).$22400 \ mL$ o

Vedclass · Accepted Answer

$12$

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(A) At $STP$,$1 \ mol$ of $O_2$ occupies $22400 \ mL$ and provides $4 \ equivalents$ of oxygen (since $O_2 + 4e^- ightarrow 2O^{2-}$).$22400 \ mL$ of $O_2$ corresponds to $4 \ equivalents$ of oxygen.$46.6 \ mL$ of $O_2$ corresponds to $\frac{4}{22400} 	imes 46.6 \approx 0.00832 \ equivalents$ of oxygen.According to the law of equivalence,the number of equivalents of metal equals the number of equivalents of oxygen.$\frac{	ext{Weight of metal}}{	ext{Equivalent weight of metal}} = 0.00832$$\frac{0.1}{E} = 0.00832$$E = \frac{0.1}{0.00832} \approx 12.02 \approx 12$.

$0.1 \ g$ of metal combines with $46.6 \ mL$ of oxygen at $STP$. The equivalent weight of the metal is:

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