The atomic weight of Fe is 56. The weight of | vedclass

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(B) The reaction for the deposition of $Fe$ from $FeCl_3$ is: $Fe^{3+} + 3e^- \rightarrow Fe(s)$.According to Faraday's law,the number of moles of met

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$11.2$

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(B) The reaction for the deposition of $Fe$ from $FeCl_3$ is: $Fe^{3+} + 3e^- ightarrow Fe(s)$.According to Faraday's law,the number of moles of metal deposited is equal to the number of Faradays passed divided by the valency factor ($n$-factor).Here,the $n$-factor for $Fe^{3+}$ is $3$.Number of moles of $Fe = \frac{	ext{Faradays passed}}{n	ext{-factor}} = \frac{0.6}{3} = 0.2 \ mol$.Weight of $Fe = 	ext{moles} 	imes 	ext{atomic weight} = 0.2 	imes 56 = 11.2 \ g$.

The atomic weight of $Fe$ is $56$. The weight of $Fe$ deposited from $FeCl_3$ solution by passing $0.6$ Faraday of electricity is $............$ $g$.

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