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Electrolysis of molten NaCl at the electrode passes $1$ F of charge. The amount of Na deposited at the cathode is:
A$23$ g, equal to the molar mass without any correction
B$0.5$ mol, half of the amount expected by Faraday's law
C$2$ mol, twice the per-mole charge correspondence here
D$1$ mol = $23$ g, since one Faraday gives one mole of $Na^+$
Answer & Solution
Correct answer: D. $1$ mol = $23$ g, since one Faraday gives one mole of $Na^+$
Na$^+ + e^- \to$ Na; 1 mole electrons (= 1 F) gives 1 mole Na = 23 g.
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