A solution contains 410.3 g of `H_2SO_4` per litre of the solution at 20°C. If its density is `1.243 g cm^(-3)`, what will be its molality and molarity ?

Answer

To solve the problem, we need to calculate both the molarity and molality of the sulfuric acid (H₂SO₄) solution.

### Step 1: Calculate the moles of H₂SO₄
We start by calculating the number of moles of H₂SO₄ using the formula:
\[ \text{Moles of solute} = \frac{\text{mass of solute (g)}}{\text{molar mass of solute (g/mol)}} \]
Given:
- Mass of H₂SO₄ = 410.3 g
- Molar mass of H₂SO₄ = 98 g/mol

Calculating the moles:
\[ \text{Moles of H₂SO₄} = \frac{410.3 \, \text{g}}{98 \, \text{g/mol}} = 4.187 \, \text{mol} \]

### Step 2: Calculate the molarity of the solution
Molarity (M) is defined as the number of moles of solute per liter of solution:
\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution (L)}} \]
Given that the volume of the solution is 1 L:
\[ \text{Molarity} = \frac{4.187 \, \text{mol}}{1 \, \text{L}} = 4.187 \, \text{M} \]

### Step 3: Calculate the mass of the solution
To find the molality, we first need to calculate the mass of the solution using its density. The density of the solution is given as 1.243 g/cm³. Since the volume of the solution is 1 L (or 1000 cm³):
\[ \text{Mass of solution} = \text{volume} \times \text{density} = 1000 \, \text{cm}^3 \times 1.243 \, \text{g/cm}^3 = 1243 \, \text{g} \]

### Step 4: Calculate the mass of the solvent (water)
Next, we find the mass of the solvent (water) by subtracting the mass of the solute (H₂SO₄) from the mass of the solution:
\[ \text{Mass of solvent} = \text{mass of solution} - \text{mass of solute} = 1243 \, \text{g} - 410.3 \, \text{g} = 832.7 \, \text{g} \]

### Step 5: Calculate the molality of the solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent:
\[ \text{Molality (m)} = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} \]
Converting the mass of the solvent from grams to kilograms:
\[ \text{Mass of solvent} = 832.7 \, \text{g} = 0.8327 \, \text{kg} \]
Now, calculating the molality:
\[ \text{Molality} = \frac{4.187 \, \text{mol}}{0.8327 \, \text{kg}} = 5.028 \, \text{m} \]

### Final Results
- Molarity of the solution: **4.187 M**
- Molality of the solution: **5.028 m**