Potassium chlorate is prepared by electrolysis of KCl in basic...

Given balanced equation is

$\overset{⊝}{60H} + \overset{⊝}{C}l ⟶ {ClO}_{3}^{-} + 3{H}_{2}O + 6{e}^{-}$

$\to 10{gKClO}_{3} ⇒ \dfrac{10}{122 . 6}{molKCO}_{3}$ in obtained

$\to$ from the above reaction, it is concluded that by
$6F$ charge $1{molKClO}_{3}$ is obtained.

$\to$ By the passage of $6F$ charge $= 1{molKClO}_{3}$

$∴$ By the passage of $\dfrac{x \times 10 \times 60 \times 60}{96500}F$ charge

$= \dfrac{1}{6} \times \dfrac{x \times 10 \times 60 \times 60}{96500}$

Now $\dfrac{x \times 10 \times 60 \times 60}{6 \times 96500} = \dfrac{10}{122 . 6}$

$⇒ x = \dfrac{10 \times 965}{60 \times 122 . 6} = \dfrac{965}{735 . 6} = 1 . 311 ≃ 1$

OR

$W = \dfrac{E}{F} \times I \times t$

$10 = \dfrac{122 . 6}{96500 \times 6} \times x \times 10 \times 3600$

$X = 1 . 311$