Question

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)

• A. 3.15 g
• B. 0.315 g
• C. 31.5 g
• D. 0.0315 g

Answer 1

The Correct Option is B

Step 1: Use the Formula for Mass of Substance Deposited

The mass of copper deposited can be calculated using the formula:

$$ m = \frac{I \cdot t \cdot M}{n \cdot F} $$

• I = 9.6487 A (current)
• t = 100 s (time)
• M = 63 g/mol (molar mass of copper)
• n = 2 (number of electrons for \( Cu^{2+} \) to \( Cu \))
• F = 96487 C/mol (Faraday constant)

Step 2: Calculate the Mass of Copper Deposited

Substitute the given values into the formula:

$$ m = \frac{9.6487 \times 100 \times 63}{2 \times 96487} $$

After performing the calculation:

$$ m = 0.315 \, \text{g} $$

Step 3: Conclusion

The correct answer is:

Option (2): 0.315 g