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 in a 0.2 m binary aqueous solution of camphor, we will follow these steps:

### Step 1: Understand the given information
We are given a 0.2 m (molality) solution of camphor (C10H16O). This means there are 0.2 moles of camphor in 1 kg of solvent (which is water in this case).

### Step 2: Calculate the number of moles of water
The molar mass of water (H2O) can be calculated as follows:
- Hydrogen (H) has an atomic mass of approximately 1 g/mol. Since there are 2 hydrogen atoms, their total mass is \(2 \times 1 = 2\) g/mol.
- Oxygen (O) has an atomic mass of approximately 16 g/mol.
- Therefore, the molar mass of water is \(2 + 16 = 18\) g/mol.

Now, we can calculate the number of moles of water in 1 kg (1000 g):
\[ \text{Number of moles of water} = \frac{\text{mass of water}}{\text{molar mass of water}} = \frac{1000 \text{ g}}{18 \text{ g/mol}} \approx 55.56 \text{ moles} \]

### Step 3: Calculate the total number of moles in the solution
We have:
- Moles of camphor = 0.2 moles
- Moles of water = approximately 55.56 moles

Total moles in the solution:
\[ \text{Total moles} = \text{moles of water} + \text{moles of camphor} = 55.56 + 0.2 = 55.76 \text{ moles} \]

### Step 4: Calculate the mole fraction of the solvent (water)
The mole fraction of the solvent (water) can be calculated using the formula:
\[ \text{Mole fraction of solvent} = \frac{\text{moles of solvent}}{\text{total moles in solution}} = \frac{\text{moles of water}}{\text{total moles}} = \frac{55.56}{55.76} \]

Calculating this gives:
\[ \text{Mole fraction of solvent} \approx 0.9964 \]

### Step 5: Conclusion
The mole fraction of the solvent (water) in the solution is approximately 0.9964.

### Final Answer:
The mole fraction of the solvent in the 0.2 m binary aqueous solution of camphor is **0.9964**.

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