If $9 g$ of $H_{2} O$ is electrolyzed completely with $50 %$ current efficiency:

1 F

of electricity will be needed

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3 F

of electricity is needed

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5.6 L

O_{2}

S T P

will be formed

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11.2 L

O_{2}

S T P

will be formed

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Solution

The correct option is C $5.6 L$ of $O_{2}$ at $S T P$ will be formed
Given weight of water $= 9 g$
Molecular weight of water $= 18 g$
As we know that,
$No. of moles = \frac{given weight}{molecular weight}$
$∴$ no. of moles of water in 9 g $= \frac{9}{18} = 0.5 moles$
Balanced equation for electrolysis of water;
$2 H_{2} O ⟶ 2 H_{2} + O_{2}$
From the reaction,
2 moles of water dissociates to produce 1 mole of oxygen at STP
$∴$ 0.5 moles of water dissociates to produce 0.25 mole of oxygen at STP
Therefore, At STP,
Volume occupied by 1 mole of oxygen $= 22.4 L$
Volume occupied by 0.25 mole of oxygen $= 0.25 \times 22.4 = 5.6 L$

Hence, option C is correct.

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