0.9 grams of acidulated water require ---- F of electricity for electrolysis

Cathode : $\left[2{\mathrm{H}}_2\mathrm{O}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2+2{\mathrm{OH}}^{-}\right]\times 2$

Anode : $2{\mathrm{H}}_2\mathrm{O}\to 4{\mathrm{H}}^{{ }^{+}}+{\mathrm{O}}_2+4{\mathrm{e}}^{-}$

Net Reaction: $2{\mathrm{H}}_2\mathrm{O}\stackrel{4\mathrm{F}}{\to }\underset{\mathrm{Cathode}}{2{\mathrm{H}}_2}+\underset{\mathrm{Anode}}{{\mathrm{O}}_2}$;

2 moles of water can be decomposed by 4 F

‘0.05’ moles (0.9 g) of water can be decomposed by ‘X’ Faradays.

‘X’ = 0.1 ;