A solution contains `410.3 g` `H_(2) SO_(4)` per litre of the solution at `20^(@)C`. If the density `= 1.243 g mL^(-1)`, what will be its molality and molarity?

Options

A. 4.187 M , 5.03 m
B. 41.87 M, 50.3 m
C. 0.4187 M, 0.503m
D. 14.87M, 50.3 m

Answer

To solve the problem, we need to calculate both the molarity and molality of the solution containing sulfuric acid (H₂SO₄). Let's break it down step by step.

### Step 1: Identify Given Data
- Mass of H₂SO₄ = 410.3 g
- Density of the solution = 1.243 g/mL
- Volume of the solution = 1 L = 1000 mL

### Step 2: Calculate the Molar Mass of H₂SO₄
The molar mass of H₂SO₄ can be calculated as follows:
- H: 1 g/mol × 2 = 2 g/mol
- S: 32 g/mol × 1 = 32 g/mol
- O: 16 g/mol × 4 = 64 g/mol

Adding these together:
\[ \text{Molar mass of H₂SO₄} = 2 + 32 + 64 = 98 \text{ g/mol} \]

### Step 3: Calculate Moles of H₂SO₄
Using the formula:
\[ \text{Moles of H₂SO₄} = \frac{\text{Mass of H₂SO₄}}{\text{Molar mass of H₂SO₄}} \]
Substituting the values:
\[ \text{Moles of H₂SO₄} = \frac{410.3 \text{ g}}{98 \text{ g/mol}} \approx 4.187 \text{ moles} \]

### Step 4: Calculate Molarity
Molarity (M) is calculated using the formula:
\[ \text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} \]
Substituting the values:
\[ \text{Molarity} = \frac{4.187 \text{ moles}}{1 \text{ L}} = 4.187 \text{ M} \]

### Step 5: Calculate Mass of the Solution
Using the density to find the mass of the solution:
\[ \text{Mass of solution} = \text{Density} \times \text{Volume} \]
Substituting the values:
\[ \text{Mass of solution} = 1.243 \text{ g/mL} \times 1000 \text{ mL} = 1243 \text{ g} \]

### Step 6: Calculate Mass of the Solvent (Water)
The mass of the solvent (water) can be calculated by subtracting the mass of H₂SO₄ from the mass of the solution:
\[ \text{Mass of solvent} = \text{Mass of solution} - \text{Mass of H₂SO₄} \]
Substituting the values:
\[ \text{Mass of solvent} = 1243 \text{ g} - 410.3 \text{ g} = 832.7 \text{ g} \]

### Step 7: Calculate Molality
Molality (m) is calculated using the formula:
\[ \text{Molality} = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} \]
First, convert the mass of the solvent from grams to kilograms:
\[ \text{Mass of solvent in kg} = \frac{832.7 \text{ g}}{1000} = 0.8327 \text{ kg} \]
Now substituting the values:
\[ \text{Molality} = \frac{4.187 \text{ moles}}{0.8327 \text{ kg}} \approx 5.03 \text{ m} \]

### Final Results
- Molarity = 4.187 M
- Molality = 5.03 m

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