How much electricity should be passed through a water votameter in order to release 22.4 liter of hydrogen at N.T.P?

1 coul

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22.4 coul

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96599 coul

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1,93,000 coul

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Solution

The correct option is D 1,93,000 coul
We know 2.016 gm of hydrogen occupies 22.4 liters at N.T.P also, to liberate 1.008 gm of hydrogen charge required is 96500 C (1 Faraday).
Hence, the electricity required to release 22.4 liter of hydrogen at N.T.P is
$= \frac{2.016}{1.008} \times 96500 = 2 \times 96500 = 193000 C$

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How much electricity should be passed through a water votameter in order to release 22.4 liter of hydrogen at N.T.P?

The correct option is D 1,93,000 coulWe know 2.016 gm of hydrogen occupies 22.4 liters at N.T.P also, to liberate 1.008 gm of hydrogen charge required is 96500 C (1 Faraday).Hence, the electricity required to release 22.4 liter of hydrogen at N.T.P is=2.0161.008×96500=2×96500=193000C