Mass in grams of copper deposited by passing 9.6487 A current through a voltmeter 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
The Correct Option is B
Solution and Explanation
To calculate the mass of copper deposited when a current of 9.6487 A is passed for 100 seconds, we use Faraday's laws of electrolysis. According to these laws, the amount of substance deposited during electrolysis is directly proportional to the total electric charge passed through the solution. The relationship is given by:
n = Q / 96500
where n is the number of moles of electrons, Q is the total electric charge in coulombs, and 96500 is the Faraday constant (approximately equal to 96487 C).
Step 1: Calculate the total charge (Q).
The charge Q is calculated using the formula:
Q = I × t
where I is the current in amperes and t is the time in seconds.
Given I = 9.6487 A and t = 100 s:
Q = 9.6487 × 100 = 964.87 C
Step 2: Calculate the number of moles of copper deposited.
The reduction of Cu²⁺ to Cu requires 2 moles of electrons per mole of copper deposited. So:
n (Cu) = 964.87 / 2 × 96487
n (Cu) = 0.005
Step 3: Calculate the mass of copper deposited.
The mass (m) is calculated using the molar mass (M), given by:
m = n × M
where M = 63 g/mol for Cu.
m = 0.005 × 63 = 0.315 g
Therefore, the mass of copper deposited is 0.315 grams.
Conclusion:
The correct answer is 0.315 g.
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