JEE MAIN - Chemistry (2003 - No. 34)
When during electrolysis of a solution of AgNO3, 9650 coulombs of charge pass through the electroplating
bath, the mass of silver deposited on the cathode will be :
10.8 g
21.6 g
108 g
1.08 g
Explanation
When $$96500$$ coulomb of electricity is passed through the electroplating bath the amount of Ag deposited $$=108g$$
$$\therefore$$ when $$9650$$ coulomb of electricity is passed deposited Ag.
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {{108} \over {96500}} \times 9650 = 10.8\,g$$
$$\therefore$$ when $$9650$$ coulomb of electricity is passed deposited Ag.
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {{108} \over {96500}} \times 9650 = 10.8\,g$$
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