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

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Number of electrons required to deposit 56 grams of Iron from molten ${FeCl}_{3}$ will be

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$6 .02 \times 10^{24}$

$6 .02 \times 10^{23}$

$1 .8 \times 10^{24}$

$1 .2 \times 10^{24}$

answer is B.

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Detailed Solution

$GEW of Iron in {FeCl}_{3} = \frac{56}{3} g$

$1 F deposits \frac{56}{3} grams of Fe$

‘X’ F depositis 56 grams of Fe

X = 3F

$∴$ Number of electrons required $= 3 \times 6 .023 \times 10^{23}$

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