1 mole heptane `(V.P = 92 mm of Hg)` is mixed with 4 mol. Octane `(V.P = 31` mm of `Hg)`, form an ideal solution. Find out the vapour pressure of solution.

Options

A. 46.2 mm of Hg
B. 40.0 mm of Hg
C. 43.2 mm of Hg
D. 38.4 mm of Hg

Answer

To find the vapor pressure of the solution formed by mixing 1 mole of heptane with 4 moles of octane, we will use Raoult's Law. Here are the steps to solve the problem:

### Step 1: Identify the Given Data
- Moles of heptane (n₁) = 1 mole
- Moles of octane (n₂) = 4 moles
- Vapor pressure of pure heptane (P₁⁰) = 92 mm Hg
- Vapor pressure of pure octane (P₂⁰) = 31 mm Hg

### Step 2: Calculate Total Moles in the Solution
Total moles (n_total) = n₁ + n₂
\[ n_{total} = 1 + 4 = 5 \text{ moles} \]

### Step 3: Calculate Mole Fractions
- Mole fraction of heptane (X₁):
\[ X_1 = \frac{n_1}{n_{total}} = \frac{1}{5} = 0.2 \]

- Mole fraction of octane (X₂):
\[ X_2 = \frac{n_2}{n_{total}} = \frac{4}{5} = 0.8 \]

### Step 4: Apply Raoult's Law
According to Raoult's Law, the vapor pressure of the solution (P_solution) can be calculated as:
\[ P_{solution} = P_1⁰ \cdot X_1 + P_2⁰ \cdot X_2 \]

Substituting the values:
\[ P_{solution} = (92 \text{ mm Hg} \cdot 0.2) + (31 \text{ mm Hg} \cdot 0.8) \]

### Step 5: Perform the Calculations
Calculating the contributions from each component:
- Contribution from heptane:
\[ 92 \cdot 0.2 = 18.4 \text{ mm Hg} \]

- Contribution from octane:
\[ 31 \cdot 0.8 = 24.8 \text{ mm Hg} \]

Now add the contributions:
\[ P_{solution} = 18.4 + 24.8 = 43.2 \text{ mm Hg} \]

### Final Answer
The vapor pressure of the solution is **43.2 mm Hg**.

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