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(C) According to Faraday's law of electrolysis,the mass of substances deposited or liberated is proportional to their equivalent weights: $\frac{W_{Ag

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$5.6$

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(C) According to Faraday's law of electrolysis,the mass of substances deposited or liberated is proportional to their equivalent weights: $\frac{W_{Ag}}{E_{Ag}} = \frac{W_{O_2}}{E_{O_2}}$Given $W_{Ag} = 0.108 \ g$ and $E_{Ag} = 108 \ g/eq$,$E_{O_2} = 8 \ g/eq$.Substituting the values: $\frac{0.108}{108} = \frac{W_{O_2}}{8}$$W_{O_2} = 10^{-3} 	imes 8 = 8 	imes 10^{-3} \ g$.At $STP$,$32 \ g$ of $O_2$ occupies $22400 \ mL$.Therefore,the volume of $8 	imes 10^{-3} \ g$ of $O_2$ is: $\frac{22400 	imes 8 	imes 10^{-3}}{32} \ mL = 5.6 \ mL$.

$A$ silver voltameter is connected in series with a water voltameter and an electric current is passed through them. At the end of electrolysis,the weight of the cathode in the silver voltameter increases by $0.108 \ g$. What is the volume of oxygen liberated at $STP$ in $mL$?

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