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JEE Main levelNumerical answerDifficulty 1 of 3ec19

The time (in s, nearest integer) needed to deposit 1.00 g of aluminium from molten $Al X_{2} O X_{3}$ with a current of 10.0 A at 100% efficiency is. (Al = 27)

Numerical answer type, answer in s, tolerance 2.

Answer: 1072 s

Al X^{3 +} + 3 e X^{-} Al

. Moles of electrons

= 3 \times \frac{1.00}{27} = 0.1111

. Charge

= 0.1111 \times 96500 = 10722 C

.

t = Q / I = 10722/10 = 1072 s

.

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