If nearly 10 $^{5}$ coulomb are liberated by 1gm equivalent of aluminium, then amount of aluminium (equivalent weight 9) deposited through electrolysis in 20 minutes by a current of 50 ampere will be

0.6 gm

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0.9 gm

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5.4 gm

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10.8 gm

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Solution

The correct option is C 5.4 gm
According to Faraday's law of electrolysis, the amount of mass deposited ,

$m = z I t$ where $z =$ electrochemical equivalent of substance , $I =$ current pass through it and $t =$ time.
Here, $z = \frac{9}{10^{5}}$

thus, $m = \frac{9}{10^{5}} \times 50 \times (20 \times 60) = 5.4$ gm

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If nearly 105 coulomb are liberated by 1gm equivalent of aluminium, then amount of aluminium (equivalent weight 9) deposited through electrolysis in 20 minutes by a current of 50 ampere will be

The correct option is C 5.4 gmAccording to Faraday's law of electrolysis, the amount of mass deposited , m=zIt where z= electrochemical equivalent of substance , I= current pass through it and t= time.Here, z=9105thus, m=9105×50×(20×60)=5.4 gm

If nearly 10 $^{5}$ coulomb are liberated by 1gm equivalent of aluminium, then amount of aluminium (equivalent weight 9) deposited through electrolysis in 20 minutes by a current of 50 ampere will be