Question

60 g of CH₃COOH dissolved in 1 dm³ solvent. What is molality of solution? (density = 1.25 g/cm³)

• A. 0.8 m
• B. 0.4 m
• C. 0.2 m
• D. 0.6 m

Hint:

Molality is calculated using the mass of the solvent, not the volume, and requires knowledge of the solvent's density.

Answer 1

The Correct Option is A

Step 1: Formula for molality.
Molality is defined as the number of moles of solute per kilogram of solvent. We use the formula: \[ m = \frac{n_{\text{solute}}}{m_{\text{solvent}}} \] Where: - \( n_{\text{solute}} = \frac{\text{mass of solute}}{\text{molar mass of solute}} \), - \( m_{\text{solvent}} \) is the mass of the solvent in kg.

Step 2: Calculation.
The mass of acetic acid is 60 g, and the molar mass of CH₃COOH is 60 g/mol. Hence, the number of moles of CH₃COOH is: \[ n_{\text{solute}} = \frac{60}{60} = 1 \, \text{mol} \] Now, we calculate the mass of the solvent. Given that the density of the solution is 1.25 g/cm³, the volume of the solution is 1 dm³ (1000 cm³). Thus, the total mass of the solution is: \[ \text{mass of solution} = 1.25 \times 1000 = 1250 \, \text{g} \] Since the mass of the solute is 60 g, the mass of the solvent is: \[ m_{\text{solvent}} = 1250 - 60 = 1190 \, \text{g} = 1.19 \, \text{kg} \] Now, molality is: \[ m = \frac{1}{1.19} \approx 0.84 \, \text{m} \quad \text{(approximately 0.8 m)} \]
Step 3: Conclusion.
The molality of the solution is approximately 0.8 m. Thus, the correct answer is (A) 0.8 m.