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 find the vapor pressure of pure n-heptane (P₀B), we can follow these steps:

### Step 1: Define Variables
Let:
- P₀A = Vapor pressure of pure n-hexane
- P₀B = Vapor pressure of pure n-heptane
- X_A = Mole fraction of n-hexane in the solution
- X_B = Mole fraction of n-heptane in the solution

### Step 2: Calculate Mole Fractions
Initially, we have:
- 1 mole of n-hexane (A)
- 3 moles of n-heptane (B)

Total moles = 1 + 3 = 4 moles

Thus, the mole fractions are:
- X_A = 1/4
- X_B = 3/4

### Step 3: Apply Raoult's Law
According to Raoult's Law, the total vapor pressure (P_total) of the solution is given by:
\[ P_{total} = P_A + P_B = P₀A \cdot X_A + P₀B \cdot X_B \]

From the problem, we know:
\[ P_{total} = 550 \, \text{mm Hg} \]

Substituting the values:
\[ 550 = P₀A \cdot \frac{1}{4} + P₀B \cdot \frac{3}{4} \]

### Step 4: Rearranging the Equation
Rearranging gives us:
\[ P₀A \cdot \frac{1}{4} + P₀B \cdot \frac{3}{4} = 550 \]

Multiplying through by 4 to eliminate the fraction:
\[ P₀A + 3P₀B = 2200 \, \text{(Equation 1)} \]

### Step 5: Adding 1 Mole of n-Heptane
When we add 1 mole of n-heptane, the new total moles become:
- 1 mole of n-hexane
- 4 moles of n-heptane

Total moles = 1 + 4 = 5 moles

New mole fractions are:
- X_A = 1/5
- X_B = 4/5

The new total vapor pressure becomes:
\[ P_{total} = P₀A \cdot X_A + P₀B \cdot X_B = P₀A \cdot \frac{1}{5} + P₀B \cdot \frac{4}{5} \]

From the problem, we know this new pressure is:
\[ P_{total} = 550 + 10 = 560 \, \text{mm Hg} \]

### Step 6: Set Up the Second Equation
Substituting into the equation gives:
\[ 560 = P₀A \cdot \frac{1}{5} + P₀B \cdot \frac{4}{5} \]

Multiplying through by 5:
\[ P₀A + 4P₀B = 2800 \, \text{(Equation 2)} \]

### Step 7: Solve the System of Equations
Now we have two equations:
1. \( P₀A + 3P₀B = 2200 \)
2. \( P₀A + 4P₀B = 2800 \)

Subtract Equation 1 from Equation 2:
\[ (P₀A + 4P₀B) - (P₀A + 3P₀B) = 2800 - 2200 \]
\[ P₀B = 600 \, \text{mm Hg} \]

### Step 8: Substitute Back to Find P₀A
Substituting \( P₀B = 600 \) into Equation 1:
\[ P₀A + 3(600) = 2200 \]
\[ P₀A + 1800 = 2200 \]
\[ P₀A = 400 \, \text{mm Hg} \]

### Conclusion
The vapor pressure of pure n-heptane (P₀B) is:
\[ P₀B = 600 \, \text{mm Hg} \]