Question

Sulphuric acid used in Lead Storage battery has a concentration of 4.5 M and a density of 1.28 g/ml. The molality of the acid is __________.

• A. 4.012
• B. 2.568
• C. 5.364
• D. 3.516

Hint:

To calculate molality, always remember that it involves the mass of the solvent in kilograms. If you have the molarity and density, use them to find the mass of the solvent before proceeding with the calculation.

Answer 1

The Correct Option is C

To calculate the molality (\(m\)) of the sulphuric acid, we can use the formula: \[ m = \frac{n_{\text{solute}}}{m_{\text{solvent}}} \] Where: - \(n_{\text{solute}}\) is the number of moles of solute, - \(m_{\text{solvent}}\) is the mass of the solvent in kilograms. We are given: - The molarity (\(M\)) of the solution is 4.5 M, which means 4.5 moles of H\(_2\)SO\(_4\) are present in 1 liter of solution. - The density (\(\rho\)) of the solution is 1.28 g/ml, which is equivalent to 1280 g/L. The mass of the solution is: \[ \text{mass of solution} = 1280 \, \text{g/L} \] The mass of the solute (H\(_2\)SO\(_4\)) is: \[ \text{mass of solute} = 4.5 \, \text{mol/L} \times 98 \, \text{g/mol} = 441 \, \text{g} \] The mass of the solvent (water) is: \[ \text{mass of solvent} = 1280 \, \text{g} - 441 \, \text{g} = 839 \, \text{g} = 0.839 \, \text{kg} \] Now, we can calculate the molality (\(m\)): \[ m = \frac{4.5 \, \text{mol}}{0.839 \, \text{kg}} = 5.364 \, \text{mol/kg} \]
Thus, the molality of the acid is 5.364 mol/kg.