Potassium chlorate is prepared by the electrolysis of KCl in basic solution. $$ 6 \mathrm{OH}^{-}+\mathrm{Cl}^{-} \rightarrow \mathrm{ClO}^{-}{ }_3+3 \mathrm{H}_2 \mathrm{O}+6 \mathrm{e}^{-} $$
If only $60 \%$ of the current is utilized in the reaction, the time (rounded to the nearest hour) required to produce 10 g of $\mathrm{KClO}_3$ using a current of 2 A is (Given : $F=96,500 \mathrm{C} \mathrm{mol}^{-1} ;$ molar mass of $\mathrm{KClO}_3=122 \mathrm{~g} \mathrm{~mol}^{-1}$ )