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Assuming no change in volume, the time required to obtain solution of $\mathrm{pH}=4$ by electrolysis of $100 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{NaOH}$ (using current $0.5 \mathrm{~A}$ ) will be
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The correct answer is:
1.93s
$\left[\mathrm{H}^{+}\right]=10^{-4} \mathrm{M}$
No. of moles of $\mathrm{NaOH}$ in $100 \mathrm{~mL}$ solution
$=\frac{M V}{1000}=\frac{10^{-4} \times 100}{1000}=10^{-5}$
Mass of $\mathrm{NaOH}$ in solution $=10^{-5} \times 40=4 \times 10^{-4} \mathrm{~g}$
By Faraday's law,
$W=\frac{I t E}{96500}$
$\begin{aligned} \text{or} \quad 4 \times 10^{-4} & =\frac{0.5 \times t \times 40}{96500} \\ t & =\frac{4 \times 10^{-4} \times 96500}{0.5 \times 40}=1.93 \mathrm{~s}\end{aligned}$
No. of moles of $\mathrm{NaOH}$ in $100 \mathrm{~mL}$ solution
$=\frac{M V}{1000}=\frac{10^{-4} \times 100}{1000}=10^{-5}$
Mass of $\mathrm{NaOH}$ in solution $=10^{-5} \times 40=4 \times 10^{-4} \mathrm{~g}$
By Faraday's law,
$W=\frac{I t E}{96500}$
$\begin{aligned} \text{or} \quad 4 \times 10^{-4} & =\frac{0.5 \times t \times 40}{96500} \\ t & =\frac{4 \times 10^{-4} \times 96500}{0.5 \times 40}=1.93 \mathrm{~s}\end{aligned}$
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