Problem 19 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Liquids A and B form ideal solution over the entire range of composition. At temperature \(T\), equimolar binary solution of liquids \(\mathbf{A}\) and \(\mathbf{B}\) has vapour pressure 45 Torr. At the same temperature, a new solution of \(\mathbf{A}\) and \(\mathbf{B}\) having mole fractions \(x_{A}\) and \(x_{B}\), respectively, has vapour pressure of \(22.5\) Torr. The value of \(x_{A} / x_{B}\) in the new solution is ____ . (given that the vapour pressure of pure liquid \(\mathbf{A}\) is 20 Torr at temperature \(T\) )

Step-by-Step Solution

Answer

The value of \( \frac{x_A}{x_B} \) in the new solution is 19.

Step 1: Understanding the Equimolar Solution

For an equimolar solution of substances A and B, the mole fractions are equal, meaning \( x_A = x_B = 0.5 \). According to Raoult's Law for ideal solutions, the total vapour pressure of the solution is given by: \[ P_{ ext{solution}} = P_A^0 x_A + P_B^0 x_B \] Here, the vapour pressure of the equimolar solution is 45 Torr.

Step 2: Finding the Vapour Pressure of Pure B

In order to find the vapour pressure of pure substance B, we need to substitute the known values into the equation for the equimolar solution: \[ 45 = (20)(0.5) + P_B^0(0.5) \] Solving for \( P_B^0 \), we have: \[ 45 = 10 + 0.5 P_B^0 \] \[ 35 = 0.5 P_B^0 \] \[ P_B^0 = 70 \, \text{Torr} \] Thus, the vapour pressure of pure liquid B is 70 Torr.

Step 3: Using Raoult's Law for the New Solution

For the new solution, apply Raoult's Law using the given overall vapour pressure of 22.5 Torr: \[ 22.5 = 20x_A + 70x_B \] Substituting the mole fraction relationship \( x_A = 1 - x_B \), we reformulate the equation: \[ 22.5 = 20(1 - x_B) + 70x_B \] \[ 22.5 = 20 - 20x_B + 70x_B \] \[ 22.5 = 20 + 50x_B \]

Step 4: Solving the Equation for Mole Fractions

Continuing from the previous step, solve for \( x_B \): \[ 22.5 - 20 = 50x_B \] \[ 2.5 = 50x_B \] \[ x_B = 0.05 \] Since \( x_A = 1 - x_B \), we'll find \( x_A \): \[ x_A = 1 - 0.05 = 0.95 \] Therefore, the expression \( \frac{x_A}{x_B} \) is: \[ \frac{0.95}{0.05} = 19 \]

Key Concepts

Raoult's Law

Raoult's Law is a fundamental principle in chemistry that helps us understand how the vapour pressure of a solution changes when we mix different substances. It states that the vapour pressure of an ideal solution is equal to the sum of the vapour pressures of its individual components, each multiplied by their respective mole fractions. In simpler terms, Raoult's Law tells us how much each component contributes to the total vapour pressure of a solution. This is important because it helps predict how mixtures behave, especially when the components are volatile, meaning they easily turn into gas. For example, if a solution is made by mixing two liquids, A and B, the vapour pressure of this solution can be calculated as:

• The vapour pressure of A when it is alone, multiplied by the mole fraction of A in the solution.
• The vapour pressure of B when it is alone, multiplied by the mole fraction of B in the solution.

Add these two results together, and you'll get the total vapour pressure of the mixture. This law is typically applicable to ideal solutions where the interactions between different molecules are similar to those in pure substances. It's not just a puzzle for chemists but a useful tool in industries where the composition of liquids and vapours is controlled, such as perfumes, gases, and even some food products. In this exercise, Raoult's Law is used to determine the vapour pressure of a new mixture by using known pressures of pure components and the conditions given.

Vapour Pressure

Vapour pressure is a key concept when discussing solutions and their behaviours. It refers to the pressure exerted by the vapour in equilibrium with its liquid or solid form. In simpler words, vapour pressure is the force that the vapour of a liquid applies when it's in a closed container or environment. When a liquid is placed in a container, some of its molecules escape into the air as gas. This gas, or vapour, will exert pressure, which is the vapour pressure. A high vapour pressure means a liquid easily evaporates, while a low vapour pressure indicates the opposite. In our example, surey A and B hold specific vapour pressures when pure, 20 Torr for A and 70 Torr for B. An important thing to remember is that the vapour pressure depends greatly on the temperature of the solution. As the temperature rises, so does the vapour pressure. This is because an increase in temperature gives liquid molecules more energy to escape into the vapour phase, leading to higher pressure. In the exercise, vapour pressures at a specific temperature help us understand how the solution components interact and change the overall vapour pressure.

Mole Fraction

Mole fraction is a fundamental way to express the concentration of a particular component within a solution. It is defined as the ratio of the number of moles of one component to the total number of moles in the solution.Mathematically, it can be expressed as:\( x_i = \frac{n_i}{n_{\text{total}}} \)where \( x_i \) is the mole fraction of component \( i \), \( n_i \) is the number of moles of component \( i \), and \( n_{\text{total}} \) is the total number of moles in the solution.Mole fraction is a unitless quantity and it is always between 0 and 1. For example, in the given exercise with an equimolar solution, both A and B have a mole fraction of 0.5, meaning they are present in equal quantities. Together, the sum of all mole fractions in a solution will always equal 1.Understanding and calculating mole fractions is crucial in many chemical applications, especially when applying Raoult's Law. It helps determine the contribution of each component to the solution properties, such as vapour pressure, boiling point, and so on. In our exercise, the mole fraction of each component helps in determining how the entire mixture behaves as a whole and how the individual properties of the components influence the overall system.