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810 grams of acidulated water is subjected to electrolysis using 9.65 $\times$ 10⁴ amperes. Time required for electrolysis is

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45 seconds

180 seconds

90 seconds

360 seconds

answer is B.

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Detailed Solution

$2 H_{2} O \overset{4 F}{\to} 2 H_{2} + O_{2};$

36 grams of water requires 4F

810 grams of water requires 90F

q = it

90 x 96500 =9.65 $\times$ 10⁴ x t

t = 90 seconds.

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