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An electric current is passed through two electrolytic cells connected in series.One containing a solution of silver nitrate and the other solution of $H_2SO_4$.What volume of oxygen measured at $25^{\large\circ}C$ and 750mm of Hg would be liberated from $H_2SO_4$ if $8\times 10^{22}$ ions of $Ag^+$ are deposited from the silver nitrate solution?

$\begin{array}{1 1}(a)\;0.89l&(b)\;0.82l\\(c)\;0.7l&(d)\;0.72l\end{array}$

1 Answer

$8\times 10^{22}$ ions of $Ag^+=\large\frac{8\times 10^{22}}{6.023\times 10^{23}}$mol
Equivalent of $O_2=\large\frac{8\times 10^{22}}{6.023\times 10^{23}}$$=0.1329g$
$\therefore 0.1329 g$ of $O_2=6.2\times .1329litre$
$\Rightarrow 0.823litre$
Hence (b) is the correct answer.
answered Feb 17, 2014 by sreemathi.v

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