When an electric current is passed through acidified water, 112ml of hydrogen gas at N.T.P was collected at the cathode in 965 seconds. The current passed in ampere is 

${\text{2H}}_{\left(\text{aq}\right)}^{\text{+}}\text{\hspace{0.17em}+\hspace{0.17em}}{\text{2e}}^{\text{-}}\stackrel{ }{\to }{\text{\hspace{0.17em}H}}_{{\text{2}}_{\left(\text{g}\right)}}$
At NTP, 22.4 L (or 22400 mL) of $H_2 = 1$mole of $H_2$
112 ml of ${\text{H}}_{\text{2}}\text{=\hspace{0.17em}}\frac{\text{112}}{\text{22400}}\text{×\hspace{0.17em}1}$
${\text{=\hspace{0.17em}0.005\hspace{0.17em}moles\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}of\hspace{0.17em}\hspace{0.17em}H}}_{\text{2}}$
${\text{moles\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}of\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}H}}_{\text{2}}\text{=}\frac{{\text{i}}_{\left(\text{A}\right)}{\text{\hspace{0.17em}t}}_{\left(\text{S}\right)}}{\text{96500c\hspace{0.17em}/mole}}\text{×\hspace{0.17em}mole\hspace{0.17em}\hspace{0.17em}ratio}$
$\text{0.005\hspace{0.17em}\hspace{0.17em}ml\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}}\frac{{\text{i}}_{\left(\text{A}\right)}{\text{\hspace{0.17em}×\hspace{0.17em}\hspace{0.17em}965}}_{\left(\text{S}\right)}}{\text{96500\hspace{0.17em}c\hspace{0.17em}/\hspace{0.17em}mole}}\text{×\hspace{0.17em}}\frac{{\text{1\hspace{0.17em}\hspace{0.17em}mole\hspace{0.17em}\hspace{0.17em}H}}_{\text{2}}}{\text{2}}$
i = 1A