When X amperes of current is passed through m | vedclass

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(A) The reaction for the deposition of aluminium is: $Al^{3+} + 3e^- \rightarrow Al(s)$.According to Faraday's law of electrolysis,the mass deposited

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$10$ ampere

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(A) The reaction for the deposition of aluminium is: $Al^{3+} + 3e^- ightarrow Al(s)$.According to Faraday's law of electrolysis,the mass deposited $w$ is given by $w = \frac{M 	imes I 	imes t}{n 	imes F}$,where $M$ is the molar mass of $Al$ $(27 \ g/mol)$,$I$ is the current in amperes,$t$ is the time in seconds $(96.5 \ s)$,$n$ is the number of electrons involved $(3)$,and $F$ is Faraday's constant $(96500 \ C/mol)$.Substituting the values: $0.09 = \frac{27 	imes I 	imes 96.5}{3 	imes 96500}$.$0.09 = \frac{27 	imes I 	imes 96.5}{289500}$.$I = \frac{0.09 	imes 289500}{27 	imes 96.5} = \frac{26055}{2605.5} = 10 \ A$.Therefore,the value of $X$ is $10$.

List $I$ (Conversion)	List $II$ (Number of Faraday required)
$A$. $1 \text{ mol of } H_2O \text{ to } O_2$	$I$. $3F$
$B$. $1 \text{ mol of } MnO_4^- \text{ to } Mn^{2+}$	$II$. $2F$
$C$. $1.5 \text{ mol of } Ca \text{ from molten } CaCl_2$	$III$. $1F$
$D$. $1 \text{ mol of } FeO \text{ to } Fe_2O_3$	$IV$. $5F$

When $X$ amperes of current is passed through molten $AlCl_3$ for $96.5$ $s$,$0.09$ $g$ of aluminium is deposited. What is the value of $X$?

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