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(B) At $NTP$,$1 \, 	ext{mole}$ of $O_2$ gas occupies $22.4 \, L$ and corresponds to $4 \, 	ext{equivalents}$ (since $2H_2O ightarrow O_2 + 4H^+ +

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$63$

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(B) At $NTP$,$1 \, 	ext{mole}$ of $O_2$ gas occupies $22.4 \, L$ and corresponds to $4 \, 	ext{equivalents}$ (since $2H_2O ightarrow O_2 + 4H^+ + 4e^-$,$n$-factor $= 4$ per mole of $O_2$).
Number of equivalents of $O_2 = \frac{11.2 \, L}{22.4 \, L/	ext{mol}} 	imes 4 = 2 \, 	ext{equivalents}$.
According to Faraday's law of electrolysis,the number of equivalents of substance deposited at the cathode is equal to the number of equivalents of substance liberated at the anode.
Therefore,equivalents of $Cu = 2$.
Mass of $Cu = 	ext{Equivalents} 	imes 	ext{Equivalent weight of } Cu$.
Equivalent weight of $Cu = \frac{	ext{Atomic mass}}{n	ext{-factor}} = \frac{63}{2} = 31.5 \, g$.
Mass of $Cu = 2 	imes 31.5 = 63 \, g$.

$A$ current is passed for $2 \, \text{hours}$ through an acidic solution,liberating $11.2 \, L$ of oxygen at $NTP$ at the anode. What will be the amount of copper deposited at the cathode by the same current when passed through a solution of copper sulphate for the same duration? .......... $g$

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