Question

Lead storage battery contains 4.25 M \( \text{H}_2\text{SO}_4 \) which has a density of 1.24 g/ml. Calculate the molality of the aqueous solution of \( \text{H}_2\text{SO}_4 \).

• A. 6.264
• B. 3.427
• C. 5.161
• D. 4.108

Hint:

To calculate molality from molarity, you need to use the density of the solution to find the mass of the solution, then subtract the mass of the solute to find the mass of the solvent.

Answer 1

The Correct Option is C

Step 1: Understand the relationship between molarity and molality.
Molarity (M) is the number of moles of solute per liter of solution, and molality (m) is the number of moles of solute per kilogram of solvent. We are given the molarity of the solution and its density, so we need to convert this to molality.

Step 2: Use the formula for molarity.
Molarity is given by:
\[ M = \frac{\text{moles of solute}}{\text{volume of solution (in L)}} \]
We are given that the molarity of \( \text{H}_2\text{SO}_4 \) is 4.25 M. This means that there are 4.25 moles of \( \text{H}_2\text{SO}_4 \) in 1 liter of solution.

Step 3: Calculate the mass of the solution.
We are also given the density of the solution as 1.24 g/ml. Since the density is mass/volume, we can calculate the mass of 1 liter (1000 ml) of solution:
\[ \text{Mass of solution} = \text{Density} \times \text{Volume} = 1.24 \, \text{g/ml} \times 1000 \, \text{ml} = 1240 \, \text{g} \]

Step 4: Find the mass of the solvent (water).
To find the mass of the solvent, we first calculate the moles of \( \text{H}_2\text{SO}_4 \) in 1 liter of solution. The molar mass of \( \text{H}_2\text{SO}_4 \) is:
\[ M_{\text{H}_2\text{SO}_4} = 2(1) + 32 + 4(16) = 98 \, \text{g/mol} \]
Thus, the mass of \( \text{H}_2\text{SO}_4 \) in 4.25 moles is:
\[ \text{Mass of } \text{H}_2\text{SO}_4 = 4.25 \, \text{mol} \times 98 \, \text{g/mol} = 416.5 \, \text{g} \]
Now, the mass of the solvent (water) is the total mass of the solution minus the mass of \( \text{H}_2\text{SO}_4 \):
\[ \text{Mass of water} = 1240 \, \text{g} - 416.5 \, \text{g} = 823.5 \, \text{g} \]
Convert this to kilograms:
\[ \text{Mass of water} = \frac{823.5 \, \text{g}}{1000} = 0.8235 \, \text{kg} \]

Step 5: Calculate the molality.
Molality is given by:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}} \]
We already know that there are 4.25 moles of \( \text{H}_2\text{SO}_4 \) in the solution and the mass of water is 0.8235 kg. So: \[ m = \frac{4.25 \, \text{mol}}{0.8235 \, \text{kg}} = 5.161 \, \text{mol/kg} \]

Step 6: Conclusion.
Thus, the molality of the solution is \( 5.161 \, \text{mol/kg} \), which corresponds to option (C).