Single Choice

If 96500 coulombs of electricity liberates one gram equivalent of anysubstance, the time taken for a current of 0.15 amperes to deposite20mg of copper from a solution of copper sulphate is (Chemicalequivalent of copper = 32)

A5 min 20 sec
B6 min 42 sec
Correct Answer
C4 min 40 sec
D5 min 50 sec

Solution

It is known that mass deposited $$ M = Z it$$
$$ \Rightarrow M= 20 \times 10 ^{-3} = \left ( \dfrac{32}{96500} \right ) \times 0.15 \times t $$
= 6.7 min = 6 min 42 sec

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