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

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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 connected in series,the masses of substances deposited are proportional to their equivalent weights. $\frac{	ext{Mass of silver}}{	ext{Mass of hydrogen}} = \frac{	ext{Eq. wt. of silver}}{	ext{Eq. wt. of hydrogen}}$ 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 $w$. $\frac{w}{5.04 	imes 10^{-2}} = \frac{108}{1.008}$ $w = \frac{108 	imes 5.04 	imes 10^{-2}}{1.008}$ $w = 100 	imes 5.04 	imes 10^{-2} = 5.4 \ g$ Thus,the mass of silver deposited is $5.4 \ g$.

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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