Q52 / 180 · COMEDK 2025 (10 May Shift 2) · Chemistry
Lead storage battery contains $4.25 \mathrm{M}$ $\mathrm{H}_2 \mathrm{SO}_4$ which has a density of $1.24 \mathrm{~g} / \mathrm{ml}$. Calculate the molality of aqueous solution of $\mathrm{H}_2 \mathrm{SO}_4$.
- A$4.108$
- B$6.264$
- Correct answerC$5.161$
- D$3.427$
Solution
Given molarity $M = 4.25 \text{ mol/L}$ and density $d = 1.24 \text{ g/mL}$.
The molar mass of $\text{H}_2\text{SO}_4$ is $2 \times 1 + 32 + 4 \times 16 = 98 \text{ g/mol}$.
Consider $1 \text{ L}$ of the solution. The mass of the solution is $1000 \text{ mL} \times 1.24 \text{ g/mL} = 1240 \text{ g}$.
The mass of the solute ($\text{H}_2\text{SO}_4$) in $1 \text{ L}$ is $4.25 \text{ mol} \times 98 \text{ g/mol} = 416.5 \text{ g}$.
The mass of the solvent (water) is $1240 \text{ g} - 416.5 \text{ g} = 823.5 \text{ g} = 0.8235 \text{ kg}$.
Molality $m$ is defined as the number of moles of solute per kilogram of solvent:
$m = \dfrac{\text{moles of solute}}{\text{mass of solvent in kg}} = \dfrac{4.25}{0.8235} \approx 5.161 \text{ mol/kg}$.
Answer: $5.161$
The molar mass of $\text{H}_2\text{SO}_4$ is $2 \times 1 + 32 + 4 \times 16 = 98 \text{ g/mol}$.
Consider $1 \text{ L}$ of the solution. The mass of the solution is $1000 \text{ mL} \times 1.24 \text{ g/mL} = 1240 \text{ g}$.
The mass of the solute ($\text{H}_2\text{SO}_4$) in $1 \text{ L}$ is $4.25 \text{ mol} \times 98 \text{ g/mol} = 416.5 \text{ g}$.
The mass of the solvent (water) is $1240 \text{ g} - 416.5 \text{ g} = 823.5 \text{ g} = 0.8235 \text{ kg}$.
Molality $m$ is defined as the number of moles of solute per kilogram of solvent:
$m = \dfrac{\text{moles of solute}}{\text{mass of solvent in kg}} = \dfrac{4.25}{0.8235} \approx 5.161 \text{ mol/kg}$.
Answer: $5.161$
Hence, the correct answer is option: 'C'.