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After electrlysis of NaCI solution with ...

After electrlysis of `NaCI` solution with inert electrodes for a certain period of time `600mL` of the solution was left. Which was found to be `1N` in `NaOH`. During the same time, `31.75 g` of `Cu` deposited in the copper voltameter in series the electrolytic cell. calculate the percentage yield of `NaOH` obtained.

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The correct Answer is:
`eta = 60%`
`Cu^(+2) +2e^(-) rarr Cu`
`n_(E) = I n_(Cu) = 0.5`
`2H_(2)O +2e^(-) rarr H_(2)+2OH^(-)`
`n_(e^(-)) = 1 n_(OH^(-)) = 1`
`N = (1)/(600) xx 100 = (5)/(3)`
%yield `=(1)/(5//3) xx 100 = 60%`
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