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JEE MAIN 2024
31.01.24_(S1)
Question
One Faraday of electricity liberates $x \times 10^{-1}$ gram atom of copper from copper sulphate, x is _____ .
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Solution
$$ \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu} $$ 2 Faraday $\rightarrow 1 \mathrm{molCu}$
1 Faraday $\rightarrow 0.5$ molCu deposit
$0.5 \mathrm{~mol}=0.5 \mathrm{~g}$ atom $=5 \times 10^{-1}$
$$ x=5 $$
Question Tags
JEE Main
Chemistry
Easy
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