Problem 121 - A Complete Resource Book in Chemistry for JEE Main

Question

Organic liquids A and B have vapour pressures \(\mathrm{p}_{1}^{0}\) and \(\mathrm{p}_{2}^{\circ}\) as pure liquids at \(80^{\circ} \mathrm{C}\). A mixture of the two liquids behaving ideally and boiling at \(80^{\circ} \mathrm{C}\) has mole fraction of \(\mathrm{A}=0.16\). If \(\left(\mathrm{p}_{2}^{\circ}-\mathrm{p}_{1}^{\circ}\right)=472 \mathrm{~mm}\) of \(\mathrm{Hg}\), what is the value of \(p_{1}^{0}\) (in \(\mathrm{mm} \mathrm{Hg}\) )? (a) \(263.6 \mathrm{~mm}\) (b) \(463.5 \mathrm{~mm}\) (c) \(663.3 \mathrm{~mm}\) (d) \(363.5 \mathrm{~mm}\)

Step-by-Step Solution

Answer

The value of \(p_1^{0}\) is 363.5 mmHg.

Step 1: Identify Given Values

We are given the mole fraction of liquid A, \( x_A = 0.16 \). We know \( (p_2^0 - p_1^0) = 472 \) mmHg. The mixture boils at \(80^{\circ}C\) and obeys Raoult's Law, indicating ideal behavior.

Step 2: Apply Raoult's Law for Boiling Point

At boiling point, the total vapor pressure equals the atmospheric pressure.For the mixture: \(P = x_A \cdot p_1^0 + x_B \cdot p_2^0\). Since the mole fraction of B, \(x_B = 1 - x_A = 0.84\), we substitute this into Raoult's Law equation.

Step 3: Set Up Equation for Total Pressure

First express \(p_2^0\) in terms of \(p_1^0\): \(p_2^0 = p_1^0 + 472 \).Substitute this into the equation: \[ P = 0.16 \cdot p_1^0 + 0.84 \cdot (p_1^0 + 472) \]Simplify this to find \(P\): \[ P = 0.16p_1^0 + 0.84p_1^0 + 396.48 = p_1^0 + 396.48 \]

Step 4: Relate Total Pressure to Atmospheric Pressure

Since the mixture boils at 80°C, we assume standard atmospheric pressure \( P = 760 \) mmHg as it boils at this temperature.Now, equate and solve for \( p_1^0 \): \[ 760 = p_1^0 + 396.48 \]

Step 5: Solve for \(p_1^0\)

Subtract 396.48 from both sides of the equation: \[ p_1^0 = 760 - 396.48 = 363.52 \] Round this value to the nearest option.

Step 6: Verify and Select Correct Answer

The calculated value \( p_1^0 \approx 363.5 \) mmHg. Verify with given options:(a) 263.6 mm(b) 463.5 mm(c) 663.3 mm(d) 363.5 mmThe correct answer is option (d) 363.5 mmHg.

Key Concepts

Vapour Pressure

Vapour pressure is a fundamental concept in physical chemistry, especially when discussing solutions and their behavior. It refers to the pressure exerted by a vapor in equilibrium with its liquid or solid form in a closed system. Whether a liquid scares its molecules off as a gas depends significantly on this pressure. The higher the vapor pressure, the more volatile the liquid is at a given temperature, which means it transforms into vapor more readily.

In the context of this exercise, we're dealing with the vapor pressures of two organic liquids, A and B. Understanding vapor pressure becomes critical when these two liquids are mixed. This is because the vapor pressure of the mixture can help determine boiling points and behaviors of solutions. When dealing with mixtures, we often apply Raoult's Law to these vapor pressures to predict behaviors like the boiling point.

Ideal Solution

An ideal solution is a concept used to simplify understanding how mixtures behave. In an ideal solution, the interactions between molecules of different components are similar to those between molecules of the same component.

It means components mix perfectly with no heat absorption or release. Each component follows Raoult's Law, which relates the partial vapor pressure of each component to its mole fraction in the solution. This type of solution obeys predictable behavior and ensures that pure components' properties can help determine the mixture's behavior.

In our exercise, the mixture of liquids A and B is assumed to behave ideally. This assumption simplifies calculations, as it lets us directly apply Raoult’s Law to find the mixture's total vapor pressure.

Boiling Point

The boiling point of a liquid is the temperature at which its vapor pressure equals the external pressure. For most practical purposes, we consider the atmospheric pressure, which is approximately 760 mm Hg at sea level.

When a solution boils, the total vapor pressure of its components equals this external pressure. Our exercise highlights that the mixture boils at 80°C. Recognizing the boiling point is essential in thermochemistry since it ties directly to a substance's physical properties like vapor pressure.

In calculations, knowing the boiling point allows us to determine essential characteristics like the vapor pressure at boiling, which helps solve problems related to Raoult's Law and ideal solutions.

Mole Fraction

Mole fraction is a way to express the concentration of a component in a mixture. It is defined as the ratio of moles of a single component to the total moles of all components in the mixture. It is dimensionless and always numbers between 0 and 1.

In the context of our exercise, the mole fraction of substance A is given as 0.16. This is a critical value because it helps determine the relative contribution of each component's vapor pressure to the total pressure using Raoult's Law.

• Mole fraction of A, denoted as \( x_A \), is 0.16.
• Mole fraction of B, \( x_B \), is derived as \( 1 - x_A \).

Using mole fractions, we effectively relate amounts in solutions to their output pressures and thereby solve for characteristics like vapor pressures and boiling points, as shown in the solution steps.