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(A) According to Faraday's law of electrolysis,the amount of substance deposited is given by $w = \frac{M 	imes Q}{n 	imes F}$,where $M$ is the mola

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$NaCl$

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(A) According to Faraday's law of electrolysis,the amount of substance deposited is given by $w = \frac{M 	imes Q}{n 	imes F}$,where $M$ is the molar mass,$Q$ is the charge in Faraday,and $n$ is the valency of the metal ion. For $1 \ g$ atom of metal,the amount deposited must be equal to its atomic mass $(M)$. Thus,$M = \frac{M 	imes 1}{n}$,which implies $n = 1$. Among the given options,$Na^+$ has a valency of $1$. Therefore,$1 \ Faraday$ of electricity will liberate $1 \ g$ atom of $Na$ from $NaCl$ solution.

From the solution of which of the following will one Faraday of electricity liberate one gram atom of metal?

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