810 grams of acidulated water is subjected to electrolysis using 9.65$\times$10⁴ amperes. Time required for electrolysis is 

$2{\mathrm{H}}_2\mathrm{O}\stackrel{4\mathrm{F}}{\to }2{\mathrm{H}}_2+{\mathrm{O}}_2;$

36 grams of water requires 4F

810 grams of water requires 90F

q = it

90 x 96500 =9.65$\times$10⁴ x t

t = 90 seconds.