$\begin{array}{1 1}(a)\;4.2l&(b)\;6.2l\\(c)\;5.3l&(d)\;3.2l\end{array}$

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1 mole of $Ag^+$=1 equivalent of $Ag^+$

$\therefore$ Equivalent of $O_2=\large\frac{22.4}{4}$$=5.6litres$ of NTP

$\therefore \large\frac{P_1\times V_1}{T_1}= \large\frac{P_2\times V_2}{T_2}$

$\large\frac{750\times V_1}{298}=\frac{760\times 5.6}{273}$

$\therefore V_1$(volume of oxygen at $25^{\large\circ}C$ and 750mm)=$\large\frac{760\times 5.6\times 298}{273\times 750}$

$\Rightarrow 6.2litres$

Hence (b) is the correct answer.

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