If the cost of electricity required to deposit 1g of Aluminium is N4.00, how much...
If the cost of electricity required to deposit 1g of Aluminium is N4.00, how much would it cost to deposit 24g of copper? [A = 27, Cu = 64]
- A) N27.02
- B) N37.02
- C) N47.02
- D) N57.02
Correct Answer: A) N27.02
Explanation
First, let's consider the relationship between the amount of deposited metal and the cost of electricity. In electrolysis, the amount of deposited metal is directly proportional to the number of moles of electrons involved, which in turn is directly proportional to the cost of electricity.
To find the cost of depositing 24g of copper, we need to determine the number of moles of electrons involved in depositing 1g of aluminum and 24g of copper. We can then use the given cost of electricity to find the cost for depositing the copper.
For aluminum (A), the relative atomic mass is 27. Therefore, 1 mole of aluminum weighs 27g. In the electrolytic process, 3 moles of electrons are required to deposit 1 mole of aluminum, as its valency is +3. Hence, for 1g of aluminum, the number of moles of electrons is:
\[\frac{1}{27} \times 3 = \frac{1}{9}\]
For copper (Cu), the relative atomic mass is 64. Therefore, 1 mole of copper weighs 64g. In the electrolytic process, 2 moles of electrons are required to deposit 1 mole of copper, as its valency is +2. Hence, for 24g of copper, the number of moles of electrons is:
\[\frac{24}{64} \times 2 = \frac{3}{4}\]Now, we need to find the cost ratio between the number of moles of electrons required to deposit 1g of aluminum and 24g of copper. The ratio is:
\[\frac{\frac{3}{4}}{\frac{1}{9}} = \frac{3}{4} \times 9 = \frac{27}{4}\]Given that the cost of electricity required to deposit 1g of aluminum is ?4.00, the cost to deposit 24g of copper is:
\[\frac{27}{4} \times 4 = ?27.00\]So, the cost to deposit 24g of copper is ?27.00, which is the correct answer,Option A.