Problem 163 - A Complete Resource Book in Chemistry for JEE Main

Question

At \(80^{\circ} \mathrm{C}\), the vapour pressure of pure liquid 'A' is 520 \(\mathrm{mm} \mathrm{Hg}\) and that of pure liquid ' \(\mathrm{B}\) ' is \(1000 \mathrm{~mm} \mathrm{Hg}\). If a mixture solution of 'A'and 'B' boils at \(80^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure, the amount of 'A' in the mixture is (1 atm = \(760 \mathrm{~mm} \mathrm{Hg}) .\) (a) \(52 \mathrm{~mol}\) per cent (b) \(34 \mathrm{~mol}\) per cent (c) \(48 \mathrm{~mol}\) per cent (d) \(50 \mathrm{~mol}\) per cent

Step-by-Step Solution

Answer

The amount of 'A' in the mixture is 50 mol per cent.

Step 1: Understanding the boiling point condition

The given problem states a mixture of liquids A and B boils at 80°C under 1 atm pressure. 1 atm pressure is equal to 760 mm Hg. To identify the mole percentage of liquid A, we need to apply Raoult's Law, which involves calculating the vapour pressure.

Step 2: Applying Raoult's Law

According to Raoult's Law, the total vapour pressure of the mixture \(P_{total}\) is the sum of the partial vapour pressures of A and B: \(P_{total} = P_A + P_B\). Each partial pressure is the product of the vapour pressure of the pure component and its mole fraction. Thus, \(P_A = x_A \times 520\) and \(P_B = x_B \times 1000\).

Step 3: Using the boiling point condition

Since the mixture boils at 1 atm pressure, \(P_{total} = 760\ mm \ Hg\). Substitute the expressions for \(P_A\) and \(P_B\) from Step 2 into this equation, we get:\[760 = x_A \times 520 + (1-x_A) \times 1000\].

Step 4: Solving the equation

Rearrange the equation from Step 3 to solve for \(x_A\):\[760 = 520x_A + 1000 - 1000x_A\]. Simplifying gives:\[-240 = -480x_A \] or \[x_A = \frac{240}{480} = \frac{1}{2}\]. Therefore, the mole fraction of A is 0.5, which corresponds to the mole percentage.

Step 5: Calculate the mole percentage of A

The mole fraction \(x_A=0.5\) translates to the mole percentage of A being \(0.5 \times 100 = 50\%\).

Key Concepts

Vapor Pressure

Vapor pressure is a fundamental concept that helps to understand how a liquid behaves under different temperatures in a closed system. It refers to the pressure exerted by the vapor of a liquid when it is in equilibrium with its liquid form.
When a liquid is in a container, some of its molecules escape into the gaseous phase. The vapor pressure is a measure of the tendency of molecules to escape from the liquid to the vapor phase.
Several factors affect vapor pressure, notably temperature and the nature of the liquid.

• **Temperature**: As temperature increases, more liquid molecules have enough kinetic energy to escape into the vapor phase, thus increasing the vapor pressure.
• **Nature of the Liquid**: Different liquids have varying intrinsic vapor pressures. For example, volatile liquids like acetone have higher vapor pressures compared to less volatile liquids like water.

Vapor pressure is vital when studying mixtures because it influences the boiling point conditions, as in the Raoult's Law application seen in the exercise.

Boiling Point

The boiling point of a liquid is the temperature at which its vapor pressure equals the external pressure exerted on it. At this stage, the liquid turns into vapor. In the context of solutions, such as a mixture of liquids A and B, the boiling point is affected by the vapor pressures of the individual components.
Raoult's Law plays a crucial role in determining the boiling point of mixtures. This principle states that the partial vapor pressure of each component in an ideal solution is directly proportional to its mole fraction.

• **Boiling Point and Mixtures**: In a mixture, the total vapor pressure is the sum of the partial pressures from each component. For a solution to boil, the total vapor pressure must equal the atmospheric pressure.
• **Impact on Calculations**: In the given exercise, the boiling condition dictates the mixture boils at 80°C with a total pressure of 760 mm Hg, matching the atmospheric pressure.

Understanding these boiling conditions helps calculate the mole fractions necessary for a specific boiling point under given constraints.

Mole Fraction

The mole fraction is a way of expressing the concentration of a component in a mixture. It is defined as the number of moles of a component divided by the total moles of all components in the mixture. This small but crucial number plays a significant role in calculating the properties of mixtures.

• **Formula**: The mole fraction of component A in a binary mixture with component B is given by: \( x_A = \frac{n_A}{n_A + n_B} \), where \( n_A \) and \( n_B \) are the moles of A and B respectively.
• **Importance in Mixtures**: Mole fraction is a dimensionless quantity that helps in expressing concentration ratios without units. It remains constant regardless of temperature and pressure changes, unlike volume-based measures.
• **Application in Raoult's Law**: The mole fraction is crucial in applying Raoult's Law to determine vapor pressures in mixtures, as seen in the exercise above. Here, it helps calculate each component's partial vapor pressure, contributing to the total pressure.
• **Interpretation**: In the exercise, the calculated mole fraction of A is 0.5, indicating that half of the mixture is made up of component A. This ultimately translates to a 50% mole percentage.

Understanding and using mole fraction effectively is key to exploring solution behaviors and performing calculations based on thermodynamic laws.