When an electric current is passed through acidulated, water 112 mL of hydrogen gas at NTP collects at the cathode in 965 seconds. The current passed, in amperes, is

22,400 mL of hydrogen at STP (or NTP) = 2gSo 112 mL of hydrogen at STP =
$\frac{2\mathrm{g}+112\mathrm{mL}}{22,400\mathrm{mL}}={10}^{-2}\mathrm{g}2{\mathrm{H}}^{+}+2{\mathrm{e}}_{2\mathrm{F}}^{-}\to {\mathrm{H}}_{1\mathrm{mol}}=2\times 96500\mathrm{C}=2\mathrm{g}$
2g hydrogen is deposited by = 2 x 96500CSo 10^-2g hydrogen will be deposited by =
$$\begin{gathered} \frac{2\mathrm{X}96500\mathrm{X}{10}^{-2}\mathrm{g}}{2\mathrm{g}} \\ =965\mathrm{C} \\ \text{Charge }\left(\mathrm{Q}\right)=\text{ it} \\ \text{So I }=\frac{\mathrm{Q}}{\mathrm{t}} \\ =\frac{965\mathrm{c}}{965\mathrm{s}} \\ =1{\mathrm{Cs}}^{-1}=1\mathrm{A} \end{gathered}$$