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If \( 9 g m H_{2} O \) is electrolysed completely with the current of \( 50 \% \) efficiency then :

1. 96500 charge is required
2. \( 2 \times 96500 C \) charge is required
3. \( 5.6 L \) of \( O_{2} \) at STP will be formed
4. \( 11.2 L \) of \( O_{2} \) at STP will be formed

Correct answer: \( 2 \times 96500 C \) charge is required

Solution

Electrolysis of water needs 2 moles of electrons per mole of H2O. For 9 g H2O, that is 0.5 mol H2O, so ideal charge is 1 mol e− = 96500 C; at 50% efficiency, the required charge doubles to 2×96500 C.

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