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(C) According to Faraday's second law of electrolysis,when the same quantity of electricity is passed through different electrolytes in series,the mas

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$5.4$

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(C) According to Faraday's second law of electrolysis,when the same quantity of electricity is passed through different electrolytes in series,the masses of substances deposited are proportional to their equivalent weights.$\frac{	ext{Mass of Ag}}{	ext{Mass of } H_2} = \frac{	ext{Eq. wt. of Ag}}{	ext{Eq. wt. of } H_2}$Given:Mass of $H_2 = 5.04 	imes 10^{-2} \ g$Eq. wt. of $H_2 = 1.008$Eq. wt. of $Ag = 108$Let the mass of silver deposited be $w$.$\frac{w}{5.04 	imes 10^{-2}} = \frac{108}{1.008}$$w = \frac{108 	imes 5.04 	imes 10^{-2}}{1.008}$$w = 107.14 	imes 5.04 	imes 10^{-2} \approx 5.4 \ g$Therefore,the mass of silver deposited is $5.4 \ g$.

List $I$ (Conversion)	List $II$ (Number of Faraday required)
$A$. $1 \text{ mole of } H_2O \text{ to } O_2$	$I$. $3F$
$B$. $1 \text{ mol of } MnO_4^- \text{ to } Mn^{2+}$	$II$. $2F$
$C$. $1.5 \text{ mol of } Ca \text{ from molten } CaCl_2$	$III$. $1F$
$D$. $1 \text{ mol of } FeO \text{ to } Fe_2O_3$	$IV$. $5F$

When the same quantity of electricity is passed through aqueous $AgNO_3$ and $H_2SO_4$ solutions connected in series,$5.04 \times 10^{-2} \ g$ of $H_2$ is liberated. What is the mass of silver (in grams) deposited? (Eq. wts. of hydrogen $= 1.008$,silver $= 108$)

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