Mass in grams of copper deposited by passing 9.6487 A current through a voltameter containing copper sulphate solution for 100 seconds is (Given : Molar mass of Cu: 63 g mol-1, 1 F = 96487 C)
- 3.15 g
- 0.315 g
- 31.5 g
- 0.0315 g
Answer
(2) 0.315 g
Solution
Q = It = 9.6487 100 = 964.87 C
Q = 964.8796487 = 0.01 F
Cu2+ + 2e- Cu: 2 F deposit 1 mol (63 g) of Cu
mass of Cu = 0.012 63
= 0.315 g
Answer: (2)