Question

$\text{Molality of 0.8 M H}_2\text{SO}_4 \text{ solution (density 1.06 g cm}^{-3}\text{) is \_\_\_\_\_\_} \times 10^{-3} \, \text{m}.$

Answer 1

Correct Answer: 815

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