$\text{Molality of 0.8 M H}_2\text{SO}_4 \text{ solution (density 1.06 g cm}^{-3}\text{) is \_\_\_\_\_\_} \times 10^{-3} \, \text{m}.$
Correct Answer: 815
Approach Solution - 1
To determine the molality of a 0.8 M H2SO4 solution with density 1.06 g cm-3, follow these steps:
1. **Understand the Definitions**: Molality (\(m\)) is the moles of solute per kilogram of solvent, while molarity (\(M\)) is the moles of solute per liter of solution.
2. **Calculate the Mass of Solution**: Given the density, 1.06 g/cm3, the mass of 1 L (1000 cm3) of solution is:
\[ 1.06 \ \text{g/cm}^3 \times 1000 \ \text{cm}^3 = 1060 \ \text{g}. \] 3. **Find Moles of Solute**: The solution is 0.8 M, meaning 0.8 moles of H2SO4 in 1 L of solution.
4. **Calculate Mass of Solute**: The molar mass of H2SO4 is 98.08 g/mol, so the mass is: \[ 0.8 \ \text{mol} \times 98.08 \ \text{g/mol} = 78.464 \ \text{g}. \] 5. **Find Mass of Solvent**: Subtract the mass of solute from the total mass of solution: \[ 1060 \ \text{g} - 78.464 \ \text{g} = 981.536 \ \text{g}. \] Convert this to kg: \[ 981.536 \ \text{g} = 0.981536 \ \text{kg}. \] 6. **Calculate Molality**: Use the formula: \[ m = \frac{\text{moles of solute}}{\text{kilograms of solvent}} = \frac{0.8 \ \text{mol}}{0.981536 \ \text{kg}} = 0.8149 \ \text{m}. \] 7. **Express in Required Form**: The molality is expressed as: \[ 0.8149 \times 10^{3} \, \text{m} \quad \text{or} \quad 814.9 \times 10^{-3} \, \text{m}. \] 8. **Verify Range**: The calculated molality (814.9) is within the given range 815,815.
Thus, the molality of the solution is approximately 815 \( \times 10^{-3} \, \text{m}\).
Approach Solution -2
To calculate the molality, we need the mass of the solvent in kilograms and the moles of \( \text{H}_2\text{SO}_4 \).
- Given molarity (\( M \)) of \( \text{H}_2\text{SO}_4 \): \( 0.8 \, \text{mol/L} \).
- Density of solution = \( 1.06 \, \text{g/cm}^3 \).
- Molar mass of \( \text{H}_2\text{SO}_4 \) = \( 98 \, \text{g/mol} \).
Step 1. Calculate the mass of 1 L of solution:
\(\text{Mass of solution} = 1.06 \times 1000 = 1060 \, \text{g}\)
Step 2. Calculate the moles of \( \text{H}_2\text{SO}_4 \) in 1 L of solution:
\(\text{Moles of } \text{H}_2\text{SO}_4 = 0.8 \, \text{mol}\)
Step 3. Calculate the mass of \( \text{H}_2\text{SO}_4 \):
\(\text{Mass of } \text{H}_2\text{SO}_4 = 0.8 \times 98 = 78.4 \, \text{g}\)
Step 4. Calculate the mass of the solvent (water) in the solution:
\(\text{Mass of water} = 1060 - 78.4 = 981.6 \, \text{g} = 0.9816 \, \text{kg}\)
Step 5. Calculate the molality (\( m \)):
\(m = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} = \frac{0.8}{0.9816} \approx 0.815 \, \text{mol/kg}\)
Step 6. Convert to \( \times 10^{-3} \) scale:
\(\text{Molality} = 815 \times 10^{-3} \, m\)
The Correct Answer is: 815
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