(D) The equivalent weight of $Al$ is calculated as $\frac{\text{Atomic weight}}{\text{Valency}} = \frac{27}{3} = 9 \ g \ eq^{-1}$.According to Faraday

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(D) The equivalent weight of $Al$ is calculated as $\frac{\text{Atomic weight}}{\text{Valency}} = \frac{27}{3} = 9 \ g \ eq^{-1}$.According to Faraday's law of electrolysis,$1 \ Faraday$ $(96500 \ C)$ of charge is required to liberate $1 \ equivalent$ of any substance.Since the mass to be liberated is $9 \ g$,which corresponds to $\frac{9 \ g}{9 \ g \ eq^{-1}} = 1 \ equivalent$,the charge required is $1 \times 96500 \ C = 96500 \ C$.

Class 12 Chemistry Electrochemistry Faraday’s law of electrolysis

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The amount of charge required to liberate $9 \ g$ of aluminium (atomic weight = $27$ and valency = $3$) in the process of electrolysis is ......... $coulombs$ (Faraday's constant $= 96500 \ C \ mol^{-1}$)

A
$321660$
B
$69500$
C
$289500$
D
$96500$

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Electrochemistry

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Faraday’s law of electrolysis

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The amount of charge required to liberate $9 \ g$ of aluminium (atomic weight = $27$ and valency = $3$) in the process of electrolysis is ......... $coulombs$ (Faraday's constant $= 96500 \ C \ mol^{-1}$)

Similar Questions

$2.5 \, F$ of electricity are passed through a $CuSO_4$ solution. The number of gram equivalents of $Cu$ deposited on the cathode is $....$

The total charge required for the complete electrolysis of $2 \ mol$ of $Na^{+}$ is ..... (in Coulombs).

What current strength in $amp$ will be required to liberate $10 \ g$ of chlorine from $NaCl$ solution in one hour?

During electrolysis of fused aluminium chloride,$0.9 \ g$ of aluminium was deposited on the cathode. The volume of chlorine liberated at the anode will be .............. $litres$ (at $STP$).

When $1 \ F$ of electricity is passed through acidulated water,the volume of $O_2$ evolved at $STP$ is ............. $dm^{3}$.

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