If 96500 coulombs of electricity liberates one gram equivalent of any substance, the time taken for a current of $0.15$ amperes to deposite $20 m g$ of copper from a solution of copper sulphate is
Given:
(Equivalent weight of copper = 32 g)

Question

If 96500 coulombs of electricity liberates one gram equivalent of any substance, the time taken for a current of

0.15

amperes to deposite

20 m g

of copper from a solution of copper sulphate is
Given:
(Equivalent weight of copper = 32 g)

Toppr Admin · Accepted Answer

Correct option is B- 6 min 42 secIt is known that mass deposited - M - Z it- -Rightarrow M-160- 20 -times-160- 10 -3- - -left - -dfrac-32-96500- -right - -times-160- 0-15 -times t -160- 6-7 min - 6 min 42 sec-160-

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)

Correct option is B. 6 min 42 sec
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

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)

Correct option is B. 6 min 42 secIt 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

Correct option is B. 6 min 42 sec
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