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

• A
  $11$
• B
  $8$
• C
  $18$
• D
  $22$