An ideal solution of two liquids is a solution in which each component obeys Raoult's which states that the vapour pressure of any component in the solution depends on the mole fraction of that component in the solution and the vapour pressure of that component in the pure state. However, there are many solution which do not obey Raoult's law. In other words, they show deviations from ideal behaviour which may be positive or negative. However, in either case, corresponding to a particular composition, they form a constant boiling mixtures called azeotropes.
A solution has a `1:4` mole ratio of pentane to hexane. The vapour presssures of the pure hydrocarbons at `20^(@)C` are 440 mm of Hg for pentane and 120 mm of Hg for hexane. The mole fraction at pentane in the vapour phase would be

Options

A. 0.2
B. 0.478
C. 0.549
D. 0.786

Explanation

To find the mole fraction of pentane in the vapor phase of a solution with a 1:4 mole ratio of pentane to hexane, we can follow these steps:

### Step 1: Determine the mole fractions of pentane and hexane in the solution.

Given the mole ratio of pentane to hexane is 1:4, we can calculate the total moles:
- Moles of pentane = 1
- Moles of hexane = 4
- Total moles = 1 + 4 = 5

Now, we can find the mole fractions:
- Mole fraction of pentane (X₁) = Moles of pentane / Total moles = 1 / 5 = 0.2
- Mole fraction of hexane (X₂) = Moles of hexane / Total moles = 4 / 5 = 0.8

### Step 2: Calculate the partial vapor pressures of pentane and hexane using Raoult's Law.

Raoult's Law states that the partial vapor pressure of each component in the solution is equal to the mole fraction of that component multiplied by its vapor pressure in the pure state.

- Vapor pressure of pure pentane (P₁°) = 440 mmHg
- Vapor pressure of pure hexane (P₂°) = 120 mmHg

Now, we can calculate the partial pressures:
- Partial pressure of pentane (P₁) = X₁ × P₁° = 0.2 × 440 mmHg = 88 mmHg
- Partial pressure of hexane (P₂) = X₂ × P₂° = 0.8 × 120 mmHg = 96 mmHg

### Step 3: Calculate the total vapor pressure of the solution.

Total vapor pressure (P_total) = P₁ + P₂ = 88 mmHg + 96 mmHg = 184 mmHg

### Step 4: Calculate the mole fraction of pentane in the vapor phase.

The mole fraction of pentane in the vapor phase (Y₁) can be calculated using the formula:
\[ Y₁ = \frac{P₁}{P_{total}} \]

Substituting the values:
\[ Y₁ = \frac{88 \text{ mmHg}}{184 \text{ mmHg}} \]

Calculating this gives:
\[ Y₁ = 0.478 \]

### Final Answer:
The mole fraction of pentane in the vapor phase is **0.478**.

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