How many grams of water will be electrolysed by $96500$ coulomb charge?

$96500C$ charge,

$H_{2}O⟶H_2+\frac{1}{2}{O}_2$

$\therefore$ mole of $e^{-}$ transferred per mole of water$=2$

$\therefore$ Chagre per mole of water $=2\times 1.6\times {10}^{-19}C\times 6.022\times {10}^{23}=19.27\times {10}^4$

$\therefore$ no. of moles of water $=\frac{96500}{19.27\times {10}^4}=0.50078$

$⟹$ No. of grams of $H_{2}O=0.50078\times 18=9.014$