When 6 g of ethanoic acid is dissolved in 200 g benzene, then what is the molality of solution ?

Question

Accepted Answer

To determine the molality of the solution when 6 g of ethanoic acid is dissolved in 200 g of benzene, we need to follow these steps:

1. **Calculate the number of moles of ethanoic acid (CH₃COOH):**
   - The molar mass of ethanoic acid is approximately 60 g/mol.
   - Number of moles = mass (g) / molar mass (g/mol) = 6 g / 60 g/mol = 0.1 mol.

2. **Calculate the mass of the solvent (benzene) in kilograms:**
   - The mass of benzene is given as 200 g, which is equivalent to 0.2 kg.

3. **Calculate the molality of the solution:**
   - Molality (m) is defined as the number of moles of solute per kilogram of solvent.
   - Molality = moles of solute / mass of solvent (kg) = 0.1 mol / 0.2 kg = 0.5 m.

Therefore, the molality of the solution is 0.5 m, which corresponds to Option 2.

**Explanation for why each of the other options is incorrect:**

- **Option 1 (0.3 m):** This option is incorrect because it underestimates the molality. The calculation clearly shows that the molality is 0.5 m, not 0.3 m.

- **Option 3 (0.6 m):** This option is incorrect because it overestimates the molality. The correct calculation yields a molality of 0.5 m, not 0.6 m.

- **Option 4 (1.0 m):** This option is incorrect because it significantly overestimates the molality. The calculated molality is 0.5 m, which is half of what is suggested by this option.

In conclusion, the correct answer is Option 2 (0.5 m), as it accurately reflects the calculated molality based on the given data.