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0.585 g of NaCl using platinum electrodes is
A. 2.0 mol
B. 0.04 mol
C. 0.1 mol
D. 0.01 mol
Ans: D
Sol: No. of mole of NaCl= 0.585/58.5= 0.01 mole
No. of mole of = 0.01 mole
No. of mole of formed = 0.01 mole