Current Electricity Ques 3
3. In producing chlorine by electrolysis $100 \mathrm{kW}$ power at $125 \mathrm{V}$ is being consumed. How much chlorine per minute is liberated? (E.C.E. of chlorine is $0.367 \times 10^{-6} $ $\mathrm{kg} / \mathrm{C}$ )
$[2010]$
(a) $1.76 \times 10^{-3} $ $\mathrm{kg}$
(b) $9.67 \times 10^{-3} $ $\mathrm{kg}$
(c) $17.61 \times 10^{-3} $ $\mathrm{kg}$
(d) $3.67 \times 10^{-3} $ $\mathrm{kg}$
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Answer:
Correct Answer: 3.(c)
Solution: (c) $\mathrm{I}=\frac{P}{V}=\frac{100 \times 10^3}{125} A=\frac{10^5}{60} A $
$ \text { E.C. E}=0.367 \times 10^{-6} \mathrm{kg} $ $C^{-1} $
$ \text { Charge per minute }=(\mathrm{I} \times 60) C $
$ =\frac{10^5 \times 60}{125} C=\frac{6 \times 10^6}{125} C $
$ \therefore \text { Mass liberated, }=\frac{6 \times 10^6}{125} \times 0.367 \times 10^{-6} $
$ =\frac{6 \times 1000 \times 0.367 \times 10^{-3}}{125} $
$ =17.616 \times 10^{-3} \mathrm{kg}$
BETA
