0.5 Faraday of electricity was passed jo deposit áll the copper present in 500ml of `CuSO_4, solution. What was the molarity of this solution?

Options

A. 1M
B. 0.5M
C. 0.25M
D. 2.5M

Answer

To find the molarity of the copper sulfate (\(CuSO_4\)) solution after passing 0.5 Faraday of electricity, we can follow these steps:

### Step 1: Understand the relation between Faraday and moles of copper
From electrochemistry, we know that:
- 2 Faraday of electricity is required to deposit 1 mole of copper (Cu).
- Therefore, 1 Faraday will deposit \( \frac{1}{2} \) mole of copper.

### Step 2: Calculate the moles of copper deposited
Given that 0.5 Faraday of electricity is passed, we can calculate the moles of copper deposited:
\[ \text{Moles of Cu deposited} = \frac{0.5 \text{ Faraday}}{2 \text{ Faraday/mole}} = 0.25 \text{ moles} \]

### Step 3: Convert the volume of the solution from mL to L
The volume of the \(CuSO_4\) solution is given as 500 mL. We need to convert this to liters:
\[ \text{Volume in L} = \frac{500 \text{ mL}}{1000} = 0.5 \text{ L} \]

### Step 4: Calculate the molarity of the solution
Molarity (M) is defined as the number of moles of solute divided by the volume of solution in liters:
\[ \text{Molarity} = \frac{\text{Moles of Cu}}{\text{Volume in L}} = \frac{0.25 \text{ moles}}{0.5 \text{ L}} = 0.5 \text{ M} \]

### Final Answer
The molarity of the \(CuSO_4\) solution is **0.5 M**.

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