Vapour pressure of solution obtained by mixing 1 mole of n hexane and 3 mole of n-heptane is 550 mm Hg . On mixing 1 mole n-heptane, vapour pressure of solution increases by 10mm Hg. Find the vapour pressure of pure n-heptane

Answer

To solve the problem, we will use Raoult's Law, which states that the vapor pressure of 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 the solution (P) = 550 mm Hg
- Moles of n-hexane (n_hexane) = 1 mole
- Moles of n-heptane (n_heptane) = 3 moles
- Increase in vapor pressure after adding 1 mole of n-heptane = 10 mm Hg

2. **Calculate Total Moles in the Initial Solution:**
\[ \text{Total moles} = n_{\text{hexane}} + n_{\text{heptane}} = 1 + 3 = 4 \text{ moles} \]

3. **Calculate Mole Fractions:**
- Mole fraction of n-hexane (X_hexane):
\[ X_{\text{hexane}} = \frac{n_{\text{hexane}}}{\text{Total moles}} = \frac{1}{4} = 0.25 \]
- Mole fraction of n-heptane (X_heptane):
\[ X_{\text{heptane}} = \frac{n_{\text{heptane}}}{\text{Total moles}} = \frac{3}{4} = 0.75 \]

4. **Apply Raoult's Law for Initial Solution:**
Using Raoult's Law:
\[ P = P^0_{\text{hexane}} \cdot X_{\text{hexane}} + P^0_{\text{heptane}} \cdot X_{\text{heptane}} \]
Substituting the known values:
\[ 550 = P^0_{\text{hexane}} \cdot 0.25 + P^0_{\text{heptane}} \cdot 0.75 \quad \text{(Equation 1)} \]

5. **Consider the New Solution After Adding 1 Mole of n-Heptane:**
- New moles of n-heptane = 3 + 1 = 4 moles
- Total moles now = 1 (n-hexane) + 4 (n-heptane) = 5 moles

6. **Calculate New Mole Fractions:**
- New mole fraction of n-hexane (X'_hexane):
\[ X'_{\text{hexane}} = \frac{1}{5} = 0.20 \]
- New mole fraction of n-heptane (X'_heptane):
\[ X'_{\text{heptane}} = \frac{4}{5} = 0.80 \]

7. **Apply Raoult's Law for New Solution:**
The new vapor pressure (P') after adding 1 mole of n-heptane is:
\[ P' = P + 10 = 550 + 10 = 560 \text{ mm Hg} \]
Using Raoult's Law again:
\[ 560 = P^0_{\text{hexane}} \cdot 0.20 + P^0_{\text{heptane}} \cdot 0.80 \quad \text{(Equation 2)} \]

8. **Solve the System of Equations:**
We now have two equations:
- Equation 1: \( 550 = P^0_{\text{hexane}} \cdot 0.25 + P^0_{\text{heptane}} \cdot 0.75 \)
- Equation 2: \( 560 = P^0_{\text{hexane}} \cdot 0.20 + P^0_{\text{heptane}} \cdot 0.80 \)

We can solve these equations simultaneously to find \( P^0_{\text{heptane}} \).

9. **Multiply Equation 1 by 0.20:**
\[ 110 = P^0_{\text{hexane}} \cdot 0.05 + P^0_{\text{heptane}} \cdot 0.15 \quad \text{(Equation 3)} \]

10. **Multiply Equation 2 by 0.25:**
\[ 140 = P^0_{\text{hexane}} \cdot 0.05 + P^0_{\text{heptane}} \cdot 0.20 \quad \text{(Equation 4)} \]

11. **Subtract Equation 3 from Equation 4:**
\[ (140 - 110) = (P^0_{\text{heptane}} \cdot 0.20 - P^0_{\text{heptane}} \cdot 0.15) \]
\[ 30 = P^0_{\text{heptane}} \cdot 0.05 \]
\[ P^0_{\text{heptane}} = \frac{30}{0.05} = 600 \text{ mm Hg} \]

### Final Answer:
The vapor pressure of pure n-heptane is **600 mm Hg**.