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

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