Potassium chlorate is prepared by electrolysis of $KCl$ in basic solution as shown by the following equation:
$6 OH^{-} + Cl^{-} \rightarrow ClO_{3}^{-} + 3 H_{2}O + 6 e^{-}$
$A$ current of $x \ A$ has to be passed for $10 \ h$ to produce $10.0 \ g$ of potassium chlorate. The value of $x$ is $.......$ (Nearest integer).
(Molar mass of $KClO_{3} = 122.6 \ g \ mol^{-1}, F = 96500 \ C \ mol^{-1}$)
$6 OH^{-} + Cl^{-} \rightarrow ClO_{3}^{-} + 3 H_{2}O + 6 e^{-}$
$A$ current of $x \ A$ has to be passed for $10 \ h$ to produce $10.0 \ g$ of potassium chlorate. The value of $x$ is $.......$ (Nearest integer).
(Molar mass of $KClO_{3} = 122.6 \ g \ mol^{-1}, F = 96500 \ C \ mol^{-1}$)
- A$1$
- B$2$
- C$3$
- D$4$
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|---|---|
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MediumKCET 2025
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