Problem 33 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The vapour pressure of ethanol and methanol are \(44.5 \mathrm{~mm}\) and \(88.7\) Hg respectively. An ideal solution is formed at the same temperature by mixing \(60 \mathrm{~g}\) of ethanol with \(40 \mathrm{~g}\) of methanol. Calculate the total vapour pressure of the solution and the mole fraction of methanol in the vapour.

Step-by-Step Solution

Answer

The total vapor pressure is 66.13 mmHg; mole fraction of methanol in vapor is 0.6567.

Step 1: Calculate the moles of ethanol and methanol

First, we need to determine the number of moles of each component in the mixture. - For ethanol (C_2H_5OH): - Molar mass of ethanol = 46 g/mol - Moles of ethanol = \( \frac{60 \, \text{g}}{46 \, \text{g/mol}} = 1.304 \, \text{mol} \)- For methanol (CH_3OH): - Molar mass of methanol = 32 g/mol - Moles of methanol = \( \frac{40 \, \text{g}}{32 \, \text{g/mol}} = 1.25 \, \text{mol} \)

Step 2: Calculate the mole fraction of each component in the liquid phase

The mole fraction is given by the ratio of moles of a component to the total moles in the mixture:- Total moles in the mixture = 1.304 mol (ethanol) + 1.25 mol (methanol) = 2.554 mol- Mole fraction of ethanol, \( x_A \) = \( \frac{1.304}{2.554} = 0.5104 \)- Mole fraction of methanol, \( x_B \) = \( \frac{1.25}{2.554} = 0.4896 \)

Step 3: Apply Raoult's Law to calculate partial pressures

Raoult's Law states that the partial pressure of each component in the solution is equal to its mole fraction in the liquid phase multiplied by its pure component vapor pressure.- Partial pressure of ethanol, \( P_A \) = \( x_A \times P^{ ext{pure}}_A = 0.5104 \times 44.5 = 22.71 \text{ mmHg} \)- Partial pressure of methanol, \( P_B \) = \( x_B \times P^{ ext{pure}}_B = 0.4896 \times 88.7 = 43.42 \text{ mmHg} \)

Step 4: Calculate the total vapor pressure of the solution

The total vapor pressure of the solution is the sum of the partial pressures of the components.- Total vapor pressure, \( P_{ ext{total}} \) = \( P_A + P_B = 22.71 + 43.42 = 66.13 \text{ mmHg} \)

Step 5: Calculate the mole fraction of methanol in the vapor

The mole fraction of methanol in the vapor, \( y_B \), can be calculated using Dalton's Law:- Mole fraction of methanol in vapor, \( y_B \) = \( \frac{P_B}{P_{ ext{total}}} = \frac{43.42}{66.13} = 0.6567 \)

Key Concepts

Vapour Pressure

Vapour pressure is an important property when dealing with solutions, especially when discussing Raoult's Law. It refers to the pressure exerted by the vapour of a liquid when it is in equilibrium with its liquid phase at a given temperature. Understanding vapour pressure helps predict how a liquid will behave in different conditions. If two components are mixed, like in the exercise, we need to look at the vapour pressures of each pure component. For example:

• Ethanol has a vapour pressure of 44.5 mmHg,
• while methanol has a vapour pressure of 88.7 mmHg.

The higher the vapour pressure, the more volatile a substance is, meaning it evaporates faster. In the exercise, the total vapour pressure is calculated based on the individual pressures of ethanol and methanol and their contributions are dependent on their mole fractions in the solution.

Ideal Solution

An ideal solution is a concept where the interactions between different molecules are similar to interactions between molecules of the same kind. This means that the molecules mix perfectly without any cavitation or expansion/contraction of volume. In such a solution, Raoult's Law is applicable, which simplifies the calculation of vapour pressure. When mixing ethanol and methanol to form an ideal solution, the solution behaves as predicted by Raoult's Law. This law assumes that the partial vapour pressure of a component is directly proportional to its mole fraction in the mixture, which makes calculations straight-forward.

Mole Fraction

The mole fraction is a way of expressing the concentration of a component in a mixture. It is calculated as the ratio of moles of one component to the total moles in the solution. In this exercise, we calculated the mole fraction for each component before proceeding with Raoult's Law.For ethanol, - Mole fraction, \( x_A \), is calculated as \( \frac{1.304}{2.554} \approx 0.5104 \). - For methanol, \( x_B \), is \( \frac{1.25}{2.554} \approx 0.4896 \).Understanding mole fractions is crucial because they help in determining how each component will contribute to the overall vapour pressure of the solution. They can be used in various contexts beyond this exercise as well, providing a clear picture of how components are distributed in any kind of mixture.

Partial Pressure

Partial pressure is the pressure exerted by a single component of a gas mixture in space. In the context of Raoult's Law, it helps us understand how much each liquid component contributes to the total pressure of the vapour above a solution.Raoult's Law tells us that the partial pressure of a component is the product of its pure component vapour pressure and its mole fraction:- For ethanol, the partial pressure \( P_A \) is \( x_A \times P^\text{pure}_A \).- For methanol, it's \( P_B = x_B \times P^\text{pure}_B \).In our exercise, we calculated each component's partial pressures and found:

• Ethanol: 22.71 mmHg,
• Methanol: 43.42 mmHg.

Adding these gives the total vapour pressure of the solution. This concept is pivotal in fields like chemistry and environmental science, where understanding contributions of individual gases is necessary.