In the aqueous solution of sulphuric acid the mole fraction of water is 0.85. The molality of the solution is-

Options

A. 8.9 m
B. 0.19 m
C. 9.8 m
D. 15 m

Answer

To solve the problem of finding the molality of a sulfuric acid solution where the mole fraction of water is 0.85, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Mole Fractions**:
- Given that the mole fraction of water (solvent) is \( x_A = 0.85 \).
- Therefore, the mole fraction of sulfuric acid (solute) is \( x_B = 1 - x_A = 1 - 0.85 = 0.15 \).

2. **Assume Total Moles**:
- Assume the total number of moles in the solution is 1. This simplifies calculations.
- Let \( n_A \) be the moles of water and \( n_B \) be the moles of sulfuric acid.
- Thus, \( n_A = 0.85 \) moles and \( n_B = 0.15 \) moles.

3. **Calculate the Moles of Solute**:
- The number of moles of solute (sulfuric acid) is \( n_B = 0.15 \) moles.

4. **Calculate the Mass of Solvent**:
- The mass of the solvent (water) can be calculated using its moles and molecular weight.
- The molecular weight of water (H₂O) is 18 g/mol.
- Therefore, the mass of water in grams is:
\[ \text{Mass of water} = n_A \times \text{Molecular weight of water} = 0.85 \times 18 = 15.3 \text{ grams} \]
- Convert this mass into kilograms:
\[ \text{Mass of water in kg} = \frac{15.3}{1000} = 0.0153 \text{ kg} \]

5. **Calculate Molality**:
- The formula for molality (m) is given by:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{n_B}{\text{mass of water in kg}} = \frac{0.15}{0.0153} \]
- Performing the calculation:
\[ m = \frac{0.15}{0.0153} \approx 9.804 \]

6. **Final Answer**:
- The molality of the sulfuric acid solution is approximately \( 9.804 \, \text{mol/kg} \).
- Rounding to significant figures, the answer is \( 9.8 \, \text{mol/kg} \).

### Conclusion:
The molality of the sulfuric acid solution is \( 9.8 \, \text{mol/kg} \).