What will be the volume of $O_{2}$ at $N T P$ liberated by $5 A$ current following for $193 s$ through acidulated water?

56 m L

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112 m L

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158 m L

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965 m L

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Solution

The correct option is A $56 m L$
According to Faraday's 1st law
$W = \frac{E Q}{96500} g$
$Q = I t$
$I = 5 A$
$t = 193 s$
$E = 8$ (equivalent weight of oxygen)
$W = \frac{8 \times 193 \times 5}{96500}$
=0.08g$.
Volume of 32g of $O_{2}$ at STP is 22400ml
So volume of 0.08g of $O_{2}$ at STP is $\frac{22400}{32} \times 0.08$
$= 56 m l$ .
Hence option A is correct.

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