Mole fraction of solute in benzene is `0.2` then what is the value of molality of solute?

Options

A. `3.2`
B. `2`
C. `4`
D. `3.6`

Explanation

To find the molality of the solute when the mole fraction of the solute in benzene is given as 0.2, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Definitions**:
- **Mole Fraction (X)**: The mole fraction of a component in a solution is defined as the ratio of the number of moles of that component to the total number of moles of all components in the solution.
- **Molality (m)**: Molality is defined as the number of moles of solute per kilogram of solvent.

2. **Set Up the Problem**:
- Let the mole fraction of the solute (X_solute) = 0.2.
- Let the mole fraction of the solvent (X_solvent) = 1 - X_solute = 1 - 0.2 = 0.8.

3. **Assume a Basis for Calculation**:
- Assume we have 1 kg (1000 g) of benzene (the solvent).
- The molar mass of benzene (C6H6) is approximately 78 g/mol.

4. **Calculate Moles of Solvent**:
- Moles of benzene (solvent) = mass of benzene / molar mass of benzene
- Moles of benzene = 1000 g / 78 g/mol ≈ 12.82 moles.

5. **Calculate Moles of Solute**:
- Using the mole fraction formula:
\[ X_{solute} = \frac{n_{solute}}{n_{solute} + n_{solvent}} \]
- Rearranging gives:
\[ n_{solute} = X_{solute} \times (n_{solute} + n_{solvent}) \]
- Substituting the values:
\[ 0.2 = \frac{n_{solute}}{n_{solute} + 12.82} \]
- Let \( n_{solute} = n \):
\[ 0.2(n + 12.82) = n \]
- Expanding and rearranging:
\[ 0.2n + 2.564 = n \]
\[ n - 0.2n = 2.564 \]
\[ 0.8n = 2.564 \]
\[ n = \frac{2.564}{0.8} \approx 3.205 \text{ moles} \]

6. **Calculate Molality**:
- Molality (m) = moles of solute / mass of solvent in kg
- Molality = \( \frac{3.205 \text{ moles}}{1 \text{ kg}} \approx 3.205 \text{ mol/kg} \).

### Final Answer:
The molality of the solute is approximately **3.205 mol/kg**.