The density of Cu is 8.94 g cm–3. The quantity of electricity needed to plate an area 10 cm x 10 cm to a thickness of 10–2 cm using CuSO4 solution would be
Answer & explanation
Correct answer: option 2
The correct answer is option 2. 27155 C.
The quantity of electricity required to plate the given area and thickness using the correct calculations.
Given:
Density of Cu = \(8.94 g cm^{-3}\)
Area = 10 cm × 10 cm = \(100 cm^2\)
Thickness = \(10^{-2} cm\)
To calculate the quantity of electricity, we need to determine the weight of copper deposited.
Weight of copper deposited = Area × Thickness × Density = \(100 cm^2 × 10^{-2} cm × 8.94 g cm^{-3}\) = \(8.94 g\)
Next, we need to convert the weight of copper to moles using its molar mass.
Molar mass of Cu = \(63.55 g/mol\)
\(\text{Moles of Cu = }\frac{Weight}{Molar mass} = \frac{8.94 g }{ 63.55 g/mol} \approx 0.1407 mol\)
The number of Faradays needed to deposit 1 mole of Cu is two.
Therefore, the quantity of electricity required can be calculated as follows:
\(\text{Quantity of electricity = Moles } \times \text{Faraday's constant = }0.1407 mol \times 2 \times 96500 C/mol \approx 27155 C\).
Therefore, the correct answer is (2) 27155 C.