At `25^(@)C`, the vapour pressure of pure methyl alcohol is 92.0 torr. Mol fraction of `CH_(3)OH` in a solution in which vapour pressure of `CH_(3)OH` is 23.0 torr at `25^(@)C`, is:

Options

A. 0.25
B. 0.75
C. 0.5 0
D. 0.66

Explanation

To solve the problem, we need to find the mole fraction of methyl alcohol (CH₃OH) in a solution where its vapor pressure is given. We will use Raoult's Law, which states that the vapor pressure of a solvent in a solution is equal to the vapor pressure of the pure solvent multiplied by its mole fraction in the solution.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Vapor pressure of pure methyl alcohol (P₀): 92.0 torr
- Vapor pressure of the solution (P): 23.0 torr

2. **Apply Raoult's Law:**
According to Raoult's Law:
\[ P = P₀ \cdot X_{CH₃OH} \]
where \(X_{CH₃OH}\) is the mole fraction of methyl alcohol in the solution.

3. **Rearranging the Equation:**
We can rearrange the equation to solve for the mole fraction:
\[ X_{CH₃OH} = \frac{P}{P₀} \]

4. **Substituting the Values:**
Now, substitute the values of P and P₀ into the equation:
\[ X_{CH₃OH} = \frac{23.0 \, \text{torr}}{92.0 \, \text{torr}} \]

5. **Calculating the Mole Fraction:**
Perform the division:
\[ X_{CH₃OH} = \frac{23.0}{92.0} = 0.25 \]

6. **Conclusion:**
The mole fraction of methyl alcohol in the solution is 0.25.

### Final Answer:
The mole fraction of methyl alcohol (CH₃OH) in the solution is **0.25**.

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