A solution containing ethyl alcohol and propyl alcohol has a vapour pressure of `290mm` at `30^(@)C`. Find the vapour pressure of pure ethyl elcohol if its mole fraction in the solution is `0.65`. The vapour pressure of propyl alcohol is `210`mm at the same temperature.

Answer

To solve the problem, we will use Raoult's Law, which states that the vapor pressure of a solution is equal to the sum of the partial vapor pressures of each component in the solution. The formula is given by:

\[ P_{total} = P^0_A \cdot X_A + P^0_B \cdot X_B \]

Where:
- \( P_{total} \) = total vapor pressure of the solution
- \( P^0_A \) = vapor pressure of pure component A (ethyl alcohol)
- \( X_A \) = mole fraction of component A (ethyl alcohol)
- \( P^0_B \) = vapor pressure of pure component B (propyl alcohol)
- \( X_B \) = mole fraction of component B (propyl alcohol)

Given:
- \( P_{total} = 290 \, \text{mmHg} \)
- \( X_A = 0.65 \)
- \( P^0_B = 210 \, \text{mmHg} \)
- \( X_B = 1 - X_A = 1 - 0.65 = 0.35 \)

We need to find \( P^0_A \) (the vapor pressure of pure ethyl alcohol).

### Step 1: Substitute the known values into the equation

Using the formula:

\[ 290 = P^0_A \cdot 0.65 + 210 \cdot 0.35 \]

### Step 2: Calculate the contribution of propyl alcohol to the total vapor pressure

Calculate \( 210 \cdot 0.35 \):

\[ 210 \cdot 0.35 = 73.5 \, \text{mmHg} \]

### Step 3: Substitute this value back into the equation

Now we have:

\[ 290 = P^0_A \cdot 0.65 + 73.5 \]

### Step 4: Isolate \( P^0_A \)

Subtract \( 73.5 \) from both sides:

\[ 290 - 73.5 = P^0_A \cdot 0.65 \]

\[ 216.5 = P^0_A \cdot 0.65 \]

### Step 5: Solve for \( P^0_A \)

Now, divide both sides by \( 0.65 \):

\[ P^0_A = \frac{216.5}{0.65} \]

Calculating this gives:

\[ P^0_A \approx 333.85 \, \text{mmHg} \]

### Final Answer

The vapor pressure of pure ethyl alcohol at 30°C is approximately **333.85 mmHg**.

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