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

Question

Organic liquids A and B have vapour pressures \(p_{1}^{\circ}\) 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 \(\mathrm{p}_{1}^{0}\) (in \(\left.\mathrm{mm} \mathrm{Hg}\right)\) ? (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 \(p_1^\circ\) is \(363.5\, \text{mm Hg}\). Therefore, the correct answer is (d).

Step 1: Understanding Raoult's Law for Ideal Mixtures

For a mixture of two components A and B, behaving ideally, Raoult's Law states that the total vapor pressure of the mixture is given by:\[ p = x_1 p_1^\circ + x_2 p_2^\circ \]where \(x_1\) and \(x_2\) are the mole fractions of components A and B, and \(p_1^\circ\) and \(p_2^\circ\) are their respective vapor pressures as pure liquids.

Step 2: Calculating Mole Fraction of B

Since the mixture is composed of only two components, A and B, the mole fraction for B \(x_2\) can be calculated as:\[ x_2 = 1 - x_1 \]Given \(x_1 = 0.16\), we have:\[ x_2 = 1 - 0.16 = 0.84 \]

Step 3: Apply Raoult's Law to Find Total Pressure

We know the mixture is boiling at \(80^\circ \text{C}\), which means the total vapor pressure \(p\) equals the external pressure at boiling point.The expression becomes:\[ p = 0.16 \cdot p_1^\circ + 0.84 \cdot p_2^\circ \]

Step 4: Using the Given Difference in Vapor Pressures

We are told that \(p_2^\circ - p_1^\circ = 472\, \text{mm Hg}\).This can be rewritten as:\[ p_2^\circ = p_1^\circ + 472 \]

Step 5: Substitute and Solve for \(p_1^\circ\)

Substitute \(p_2^\circ = p_1^\circ + 472\) into the equation from Step 3:\[ p = 0.16 \cdot p_1^\circ + 0.84 \cdot (p_1^\circ + 472) \]This simplifies to:\[ p = p_1^\circ(0.16 + 0.84) + 0.84 \cdot 472 \]\[ p = p_1^\circ + 396.48 \]Since the mixture is boiling, \( p = 760\, \text{mm Hg} \) (normal atmospheric pressure).Thus, the equation becomes:\[ 760 = p_1^\circ + 396.48 \]

Step 6: Solve for \(p_1^\circ\)

Isolate \(p_1^\circ\) in the equation:\[ p_1^\circ = 760 - 396.48 \]\[ p_1^\circ = 363.52 \]Thus, \(p_1^\circ\) rounds to \(363.5\, \text{mm Hg}\).

Key Concepts

Vapor Pressure

Vapor pressure is an essential concept in chemistry and physics that describes the pressure exerted by the vapor of a liquid or solid in equilibrium with its liquid or solid phase at a given temperature. When a liquid is placed in a closed container, some of the molecules escape from the liquid phase into the gas phase, forming vapor. This vapor exerts pressure on the walls of the container, known as vapor pressure.
A few key points about vapor pressure include:

• Vapor pressure is temperature-dependent: As the temperature increases, so does the vapor pressure because more molecules have sufficient energy to escape into the vapor phase.
• Vapor pressure is a critical factor in determining the boiling point of a liquid: A liquid boils when its vapor pressure equals the external atmospheric pressure.
• Each pure substance has a unique vapor pressure at a given temperature, making it a useful property for identifying or characterizing substances.

Understanding vapor pressure helps explain why mixtures boil at different temperatures than pure substances.

Mole Fraction

The mole fraction is a way of expressing the concentration of a component in a mixture. It is calculated by dividing the number of moles of a specific component by the total number of moles of all components in the mixture. For a two-component mixture, like the one in our exercise involving liquids A and B, the mole fraction of component A, denoted as \(x_1\), is given by:

• \(x_1 = \frac{n_1}{n_1 + n_2}\)

where \(n_1\) and \(n_2\) are the moles of components A and B, respectively.
The mole fraction has no units and always lies between 0 and 1. It is particularly significant in vapor pressure calculations because Raoult's Law uses mole fractions to determine the contribution of each component to the total vapor pressure. In the exercise, we calculate the mole fraction of B, \(x_2\), by subtracting the mole fraction of A from 1 (\(x_2 = 1 - x_1\)). This highlights the simple yet powerful utility of the mole fraction in determining component proportions in mixtures.

Ideal Mixture Behavior

Ideal mixtures are theoretical mixtures where the interactions between different molecules are identical to the interactions between molecules of the same kind. In other words, an ideal mixture behaves as if each component only "sees" the same type of molecule throughout the mixture, leading to predictable behavior based on the properties of the pure components alone.
Key characteristics of ideal mixtures include:

• Conformance to Raoult's Law: The total vapor pressure of an ideal mixture is the sum of the partial pressures of each component, each weighted by its mole fraction. This is expressed as \(p = x_1 p_1^\circ + x_2 p_2^\circ\).
• Range of similarities: Ideal behavior is more likely when the components of the mixture are chemically similar or when the temperature and pressure conditions are mild.
• Lack of interaction discrepancy: In ideal mixtures, no significant energy change is associated with mixing, meaning the components do not interact with each other more or less strongly than with themselves.

Recognizing whether a mixture can be considered ideal is crucial because it simplifies calculations and predictions about vapor pressure, boiling points, and other thermodynamic properties. In practical scenarios, many real-life mixtures exhibit nearly ideal behavior under specific conditions, making this a useful approximation.