Calculate the weights of solutes required to prepre one litre each of (a) 0.45 M `KMnO_4` solution and (b) `0.25 M CuSO_4` solution.

Answer

To solve the problem of calculating the weights of solutes required to prepare one liter each of (a) 0.45 M KMnO4 solution and (b) 0.25 M CuSO4 solution, we will follow these steps:

### Step 1: Understand the formula
The weight of the solute can be calculated using the formula:
\[ \text{Weight of solute (g)} = \text{Molarity (M)} \times \text{Molar mass (g/mol)} \times \text{Volume (L)} \]

### Step 2: Calculate for KMnO4
1. **Identify the molarity and volume**:
- Molarity (M) = 0.45 M
- Volume (L) = 1 L

2. **Find the molar mass of KMnO4**:
- The molar mass of KMnO4 is approximately 158 g/mol.

3. **Calculate the weight of KMnO4**:
\[ \text{Weight of KMnO4} = 0.45 \, \text{M} \times 158 \, \text{g/mol} \times 1 \, \text{L} = 71.1 \, \text{g} \]

### Step 3: Calculate for CuSO4
1. **Identify the molarity and volume**:
- Molarity (M) = 0.25 M
- Volume (L) = 1 L

2. **Find the molar mass of CuSO4**:
- The molar mass of CuSO4 is approximately 159.5 g/mol.

3. **Calculate the weight of CuSO4**:
\[ \text{Weight of CuSO4} = 0.25 \, \text{M} \times 159.5 \, \text{g/mol} \times 1 \, \text{L} = 39.875 \, \text{g} \]

### Final Answers:
- (a) The weight of KMnO4 required is **71.1 g**.
- (b) The weight of CuSO4 required is **39.875 g**.

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