MCQ
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 $KCIO _{3}$ using a current of $2\, A$ is..........

(Given : $F =96,500\, C\, mol ^{-1}$ molar mass of $\left. KClO _{3}=122\,gmol ^{-1}\right)$

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

Answer

Correct option: A.
$11$
a
$\frac{10}{122} \times 6=\frac{2 \times( hr ) \times 3600 \times 60 \%}{96500}$

$+( hr )=\frac{96500}{122 \times 72}=10.98 hr$

$=11$ hours

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