Electrolysis of 600 mL aqueous solution of NaCl for 5 min changes the pH of the solution to 12 .

The current in Amperes used for the given electrolysis is ___________ . (Nearest integer).

<p>Electrolysis of NaCl is</p> <p>$$\mathrm{NaCl}+\mathrm{H}_2 \mathrm{O}(\mathrm{aq}) \rightarrow \mathrm{NaOH}(\mathrm{aq})+\frac{1}{2} \mathrm{Cl}_2(\mathrm{~g})+\frac{1}{2} \mathrm{H}_2(\mathrm{~g})$$</p> <p>Since during electrolysis pH changes to 12</p> <p>So $\left[\mathrm{OH}^{\ominus}\right]=10^{-2}$ and $\left[\mathrm{H}^{+}\right]=10^{-12}$</p> <p>So by Faraday law</p> <p>Gram amount of substance deposited $=$ Amount of electricity passed</p> <p>$$\begin{aligned} & 10^{-2} \times \frac{600}{1000} \times 96500=\mathrm{I} \times \mathrm{t} \\ & \frac{10^{-2} \times 600}{1000} \times 96500=\mathrm{I} \times 5 \times 60 \\ & \mathrm{I}=\frac{10^{-2} \times 600 \times 96500}{1000 \times 5 \times 60} \\ & \mathrm{I}=1.93 \text { ampere } \end{aligned}$$</p> <p>So, $\mathrm{I = 2}$ ampere (nearest integer)</p>