What is the molarity and molality of a 13% solution (by weight) of sulphric acid with a density of `1.02 mL^(-1)`?

Explanation

To find the molarity and molality of a 13% solution (by weight) of sulfuric acid with a density of 1.02 g/mL, we will follow these steps:

### Step 1: Calculate Molarity

1. **Understand the formula for molarity**:
\[ \text{Molarity (M)} = \frac{\text{Percentage by weight} \times 10 \times \text{Density (g/mL)}}{\text{Molecular weight (g/mol)}} \]

2. **Identify the given values**:
- Percentage by weight of sulfuric acid (H₂SO₄) = 13%
- Density of the solution = 1.02 g/mL
- Molecular weight of sulfuric acid (H₂SO₄) = 98 g/mol

3. **Plug in the values into the formula**:
\[ \text{Molarity} = \frac{13 \times 10 \times 1.02}{98} \]

4. **Calculate the molarity**:
\[ \text{Molarity} = \frac{132.6}{98} \approx 1.35 \, \text{M} \]

### Step 2: Calculate Molality

1. **Understand the formula for molality**:
\[ \text{Molality (m)} = \frac{\text{Weight of solute (g)} \times 1000}{\text{Molecular weight (g/mol)} \times \text{Weight of solvent (g)}} \]

2. **Determine the weight of solute and solvent**:
- In a 100 g solution, the weight of solute (sulfuric acid) = 13 g
- The weight of solvent (water) = 100 g - 13 g = 87 g

3. **Plug in the values into the formula**:
\[ \text{Molality} = \frac{13 \times 1000}{98 \times 87} \]

4. **Calculate the molality**:
\[ \text{Molality} = \frac{13000}{8526} \approx 1.52 \, \text{m} \]

### Final Answers
- Molarity = 1.35 M
- Molality = 1.52 m