In the electrolysis of acidulated water $2.8 L$ of $O$ $_{2}$ is liberated at $S . T . P$ .
If the same quantity of electricity liberates $6 g$ of another element, the equivalent weight of the element is:

6 g / e q u i v

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12 g / e q u i v

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24 g / e q u i v

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48 g / e q u i v

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Solution

The correct option is B $12 g / e q u i v$
$22.4 L$ occupied by $1$ mole
$∴$ moles of $O_{2}$ in $2.8 L = \frac{2.8}{22.4}$

$∴$ no. of equivalent of $O_{2} = \frac{2.8 \times 4}{22.4} = 0.5$

No. of equivalents of $e^{-} = 0.5$
We know,
No. of equivalents $= \frac{w e i g h t}{e q u i v a l e n t w e i g h t}$
$0.5 = \frac{6}{E . W}$
$∴$ Equivalent weight $= 12$ g/equiv

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In the electrolysis of acidulated water 2.8L of O2 is liberated at S.T.P. If the same quantity of electricity liberates 6g of another element, the equivalent weight of the element is:

The correct option is B 12 g/equiv22.4 L occupied by 1 mole∴ moles of O2 in 2.8 L=2.822.4∴ no. of equivalent of O2=2.8×422.4=0.5No. of equivalents of e−=0.5We know,No. of equivalents =weightequivalent weight0.5=6E.W∴ Equivalent weight =12 g/equiv

In the electrolysis of acidulated water $2.8 L$ of $O$ $_{2}$ is liberated at $S . T . P$ .
If the same quantity of electricity liberates $6 g$ of another element, the equivalent weight of the element is: