A 200 W, 100 V bulb is connected in series | vedclass

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(C) The current $i$ flowing through the circuit is calculated as $i = \frac{P}{V} = \frac{200 \ W}{100 \ V} = 2 \ A$. Using Faraday's law of electroly

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$SnCl_4$

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(C) The current $i$ flowing through the circuit is calculated as $i = \frac{P}{V} = \frac{200 \ W}{100 \ V} = 2 \ A$. Using Faraday's law of electrolysis,$w = \frac{M 	imes i 	imes t}{n 	imes F}$,where $w = 11.1 \ g$,$M = 118.7 \ g \ mol^{-1}$,$i = 2 \ A$,$t = 5 	imes 3600 \ s$,and $F = 96500 \ C \ mol^{-1}$. Rearranging for $n$: $n = \frac{M 	imes i 	imes t}{w 	imes F} = \frac{118.7 	imes 2 	imes 18000}{11.1 	imes 96500} \approx 3.99 \approx 4$. Since the valency $n$ of $Sn$ is $4$,the compound is $SnCl_4$.

$A$ $200 \ W, 100 \ V$ bulb is connected in series with an electrolytic cell. If an aqueous solution of an $Sn$-salt is electrolysed for $5 \ hrs$,$11.1 \ g$ of $Sn$ gets deposited. The chemical formula of the compound is (Given atomic weight of $Sn$ is $118.7 \ g \ mol^{-1}$).

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