Problem 71 - A Complete Resource Book in Chemistry for JEE Main

Question

A solution has a 1: 4 mole ratio of pentane to hexane. The vapour pressures of the pure hydrocarbons at \(20^{\circ} \mathrm{C}\) are \(400 \mathrm{~mm} \mathrm{Hg}\) for pentane and \(120 \mathrm{~mm} \mathrm{Hg}\) for hexane. The mole fraction of pentane in the vapour phase would be (a) \(0.200\) (b) \(0.549\) (c) \(0.786\) (d) \(0.478\)

Step-by-Step Solution

Answer

None of the provided options match the calculated result of the mole fraction being approximately 0.455.

Step 1: Determine Mole Fractions in Solution

The mole ratio of pentane to hexane is given as 1:4. This means there are 1 mole of pentane and 4 moles of hexane. Calculate the total moles in the solution:\[\text{Total moles} = 1 + 4 = 5\]Next, calculate the mole fraction of pentane \(X_{C_5H_{12}}\) and hexane \(X_{C_6H_{14}}\):\[X_{C_5H_{12}} = \frac{1}{5} = 0.2\]\[X_{C_6H_{14}} = \frac{4}{5} = 0.8\]

Step 2: Apply Raoult's Law to Find Partial Pressures

Apply Raoult's Law to find the partial pressures of each component in the solution. Raoult's Law states \(P_i = X_i \cdot P_i^0\), where \(P_i\) is the partial pressure, \(X_i\) is the mole fraction, and \(P_i^0\) is the vapor pressure of the pure component.For pentane:\[P_{C_5H_{12}} = 0.2 \times 400 = 80 \, \text{mm Hg}\]For hexane:\[P_{C_6H_{14}} = 0.8 \times 120 = 96 \, \text{mm Hg}\]

Step 3: Calculate Total Pressure of the Vapour

Add the partial pressures of pentane and hexane to find the total pressure of the vapor phase:\[P_{\text{total}} = P_{C_5H_{12}} + P_{C_6H_{14}} = 80 + 96 = 176 \, \text{mm Hg}\]

Step 4: Find Mole Fraction of Pentane in Vapour Phase

The mole fraction of pentane in the vapor phase is given by dividing its partial pressure by the total pressure:\[Y_{C_5H_{12}} = \frac{P_{C_5H_{12}}}{P_{\text{total}}} = \frac{80}{176} \approx 0.455\]However, it seems there might have been a misstep as none of the options match. Let's double-check calculations and reevaluate to find the exact fraction.

Step 5: Check for Errors and Adjust Calculation

Re-evaluate the mole fraction calculation in the vapor phase since initial results did not match the expected options. Correct computing for calculation:\[Y_{C_5H_{12}} = \frac{80}{176} \approx 0.455 \] which is confirmed through recalculated methods. The process seems to need reassessment for comparison to the options and ensuring correct extraction considering errors do exist across steps.

Key Concepts

Mole Fraction

The mole fraction is a way of expressing the composition of a solution. In simple words, it tells us how much of a particular component is present in the mixture compared to the total number of moles of all components.
For example, in a solution of pentane and hexane, the mole ratio is 1:4. This means there is 1 mole of pentane and 4 moles of hexane. The total moles in the solution are calculated as follows:

• Total moles = 1 + 4 = 5

The mole fraction for pentane - which is the moles of pentane divided by the total moles - is:

• \[X_{C_5H_{12}} = \frac{1}{5} = 0.2\]

Similarly, the mole fraction for hexane is:

• \[X_{C_6H_{14}} = \frac{4}{5} = 0.8\]

These mole fractions are pivotal in calculating other properties of the solution, such as vapor pressures.

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid form at a given temperature. Every liquid has its characteristic vapor pressure, which depends on the temperature and the nature of the liquid.
In our case, pure pentane and hexane have vapor pressures of 400 mm Hg and 120 mm Hg at 20°C, respectively. When we mix these two components, their individual contributions to the total vapor pressure depend on their mole fractions and individual vapor pressures. This is where Raoult's Law comes into play, allowing us to calculate the partial pressures for each component:

• Pentane: \[ P_{C_5H_{12}} = X_{C_5H_{12}} \cdot P^0_{C_5H_{12}} = 0.2 \times 400 = 80 \, \text{mm Hg} \]
• Hexane: \[ P_{C_6H_{14}} = X_{C_6H_{14}} \cdot P^0_{C_6H_{14}} = 0.8 \times 120 = 96 \, \text{mm Hg} \]

These partial pressures help us in calculating the total pressure of the vapor phase.

Partial Pressure

Partial pressure is the pressure exerted by a single component of a mixture of gases. In our exercise, the solution of pentane and hexane results in individual partial pressures for each component in the vapor phase.
The partial pressure can be calculated using Raoult’s Law. Once we determine the partial pressures for pentane and hexane (80 mm Hg and 96 mm Hg, respectively), we add them to find the total pressure of the system:

• \[P_{\text{total}} = P_{C_5H_{12}} + P_{C_6H_{14}} = 80 \, \text{mm Hg} + 96 \, \text{mm Hg} = 176 \, \text{mm Hg} \]

Knowing partial pressures is essential because they influence the composition of the vapor above a liquid. They also determine how the mixture behaves in processes like boiling and condensation.

Mole Ratio

Mole ratio is a simple and yet fundamental concept, representing the ratio of the number of moles of one component to the number of moles of another in a mixture. It is incredibly useful in stoichiometry and in solutions.
In the given exercise, we have a 1:4 mole ratio of pentane to hexane. This directly informs us that for every 1 mole of pentane, there are 4 moles of hexane.
Understanding the mole ratio is crucial because:

• It helps in calculating mole fractions, which are necessary for determining partial pressures.
• It provides insight into the composition of mixtures, allowing scientists and students to predict how different mixtures will behave under various conditions.

This ratio ensures that calculations for reactions or processes involving the mixture are accurate and based on the proper proportions.