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Number of electrons required to deposit 56 grams of Iron from molten FeCl3 will be

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answer is 2.

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Detailed Solution

GEW of Iron in FeCl3 = 563g1 F deposits 563 grams of Fe‘X’ F depositis 56 grams of FeX = 3F∴Number of electrons required =3×6.023×1023

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