Mole fraction of solvent in 0.2 m binary aqueous solution of camphor (m = molality) is :

Options

A. 0.996
B. 0.004
C. 0.96
D. 0.976

Answer

To find the mole fraction of the solvent (water) in a 0.2 m binary aqueous solution of camphor, we can follow these steps:

### Step 1: Understand the Definitions
- **Molality (m)** is defined as the number of moles of solute per kilogram of solvent.
- **Mole fraction (X)** of a component is defined as the number of moles of that component divided by the total number of moles of all components in the solution.

### Step 2: Identify the Given Data
- We have a binary solution of camphor (solute) and water (solvent).
- The molality of the solution is given as 0.2 m.

### Step 3: Calculate Moles of Solute
- Since molality (m) = moles of solute / mass of solvent (kg), we can assume:
- Moles of camphor (solute) = 0.2 moles (as we can take 1 kg of solvent for simplicity).

### Step 4: Calculate Mass of Solvent
- We are assuming 1 kg of water (solvent), which is equivalent to 1000 grams.

### Step 5: Calculate Moles of Solvent
- The molar mass of water (H2O) is approximately 18 g/mol.
- Moles of water = mass of water (g) / molar mass of water (g/mol)

\[ \text{Moles of water} = \frac{1000 \text{ g}}{18 \text{ g/mol}} \approx 55.56 \text{ moles} \]

### Step 6: Calculate Total Moles in the Solution
- Total moles = moles of solute (camphor) + moles of solvent (water)

\[ \text{Total moles} = 0.2 + 55.56 \approx 55.76 \text{ moles} \]

### Step 7: Calculate Mole Fraction of Solvent
- The mole fraction of the solvent (water) is given by:

\[ X_{\text{water}} = \frac{\text{moles of water}}{\text{total moles}} = \frac{55.56}{55.76} \approx 0.996 \]

### Final Answer
The mole fraction of the solvent (water) in the 0.2 m binary aqueous solution of camphor is approximately **0.996**.

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