Chemistry · 2022
JEE Main · 25 June 2022, Shift 2 · Q85
A solution of Fe_2( SO_4)_3 is electrolyzed for ' x ' min with a current of 1.5 A to deposit 0.3482 g of Fe. The value of x is \ \ \ \. [nearest…
A solution of $\displaystyle \mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3$ is electrolyzed for ' x ' min with a current of $\displaystyle 1.5$ A to deposit $\displaystyle 0.3482$ g of Fe . The value of x is \_\_\_\_. [nearest integer]Given : $\displaystyle 1 \mathrm{~F}=96500 \mathrm{C} \mathrm{mol}^{-1}$
Atomic mass of $\displaystyle \mathrm{Fe}=56 \mathrm{~g} \mathrm{~mol}^{-1}$
Official answer
20
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JEE Main 2022 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.