(C) According to Faraday's law of electrolysis,the mass $m$ deposited is given by $m = \frac{E \times I \times t}{F}$,where $E$ is the equivalent weig

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(C) According to Faraday's law of electrolysis,the mass $m$ deposited is given by $m = \frac{E \times I \times t}{F}$,where $E$ is the equivalent weight,$I$ is current in amperes,$t$ is time in seconds,and $F$ is the charge required for $1 \ gm$ equivalent (given as $10^5 \ C$).Given: $E = 9$,$I = 50 \ A$,$t = 20 \ minutes = 20 \times 60 \ s = 1200 \ s$,and $F = 10^5 \ C$.Substituting the values: $m = \frac{9 \times 50 \times 1200}{10^5} = \frac{540000}{100000} = 5.4 \ gm$.

Class 12 Chemistry Electrochemistry Faraday’s law of electrolysis

Medium

If nearly $10^5 \ C$ liberate $1 \ gm$ equivalent of aluminium,then the amount of aluminium (equivalent weight $9$) deposited through electrolysis in $20 \ minutes$ by a current of $50 \ A$ will be .............. $gm$.

AIPMT 1998 All AIPMT

A
$0.6$
B
$0.09$
C
$5.4$
D
$10.8$

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Electrochemistry

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Faraday’s law of electrolysis

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AIPMT 1998 Paper All AIPMT years →

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If nearly $10^5 \ C$ liberate $1 \ gm$ equivalent of aluminium,then the amount of aluminium (equivalent weight $9$) deposited through electrolysis in $20 \ minutes$ by a current of $50 \ A$ will be .............. $gm$.

Similar Questions

To gain $1 \ mole$ $Mg$ from $Mg^{2+}$ and $1 \ mole$ $Al$ from $Al^{3+}$,how many Coulombs of electricity are required respectively?

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$A$ solution of $CuSO_4$ is electrolysed for $10$ minutes with a current of $1.5$ amperes. What is the mass of copper deposited at the cathode (in $g$)?

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