Problem 157 - A Complete Resource Book in Chemistry for JEE Main

Question

Benzene and toluene form nearly ideal solutions. At \(20^{\circ} \mathrm{C}\), the vapour pressure of benzene is 75 torr and that of toluene is 22 torr. The partial vapour pressure of benzene at \(20^{\circ} \mathrm{C}\) for a solution containing \(78 \mathrm{~g}\) of benzene and \(46 \mathrm{~g}\) of toluene in torr is (a) 25 (b) 50 (c) \(37.5\) (d) \(53.5\)

Step-by-Step Solution

Answer

The partial vapor pressure of benzene is 50 torr (option b).

Step 1: Calculate Moles of Benzene and Toluene

First, find the molar mass of benzene (C₆H₆) which is approximately 78 g/mol, and for toluene (C₇H₈), it is approximately 92 g/mol. Calculate the moles of each: moles of benzene = \( \frac{78 \, \text{g}}{78 \, \text{g/mol}} = 1\, \text{mol} \), moles of toluene = \( \frac{46 \, \text{g}}{92 \, \text{g/mol}} \approx 0.5\, \text{mol} \).

Step 2: Calculate Mole Fraction of Benzene

Mole fraction of benzene, \( x_B \), is given by \( x_B = \frac{\text{moles of benzene}}{\text{moles of benzene} + \text{moles of toluene}} = \frac{1}{1 + 0.5} = \frac{2}{3}\).

Step 3: Apply Raoult’s Law for Partial Pressure of Benzene

According to Raoult's Law, the partial pressure of benzene is \( P_B = x_B \times P_B^0 \), where \( P_B^0 \) is the vapour pressure of pure benzene (75 torr). Thus, \( P_B = \frac{2}{3} \times 75 = 50 \text{ torr} \).

Key Concepts

Ideal Solutions

In chemistry, an ideal solution is a solution that perfectly follows Raoult's Law. This means that the chemical interactions between solute and solvent are similar to those in the pure substances themselves.
In an ideal solution, the total vapour pressure of the solution is directly proportional to the mole fraction of each component and their respective pure vapour pressures. No volume change or heat exchange occurs upon mixing.
Ideal solutions are rare in real life but are a useful simplification for understanding solution behavior and properties in thermodynamics. Benzene and toluene are often used as classic examples of ideal solutions due to their similar molecular structures and interactions.

Vapour Pressure

Vapour pressure is a measure of how readily a substance moves into the gaseous phase. It's the pressure exerted by a vapour in thermodynamic equilibrium with its condensed phases at a given temperature.

• For benzene and toluene, the vapour pressure changes with temperature, but at 20°C, it is 75 torr for pure benzene and 22 torr for pure toluene.
• This concept is critical in determining boiling points and evaporation rates of substances.

Vapour pressure is crucial for understanding Raoult's Law, as it involves computing how the pressure of each component in a mixture contributes to the total vapour pressure.

Mole Fraction

The mole fraction is a way of expressing the concentration of a component in a mixture. It's calculated by dividing the moles of one component by the total moles of all components in the mixture.
For instance, in our example:

• Moles of benzene: 1 mol
• Moles of toluene: 0.5 mol

Total moles in the solution = 1 mol + 0.5 mol = 1.5 mol
The mole fraction of benzene is then \( \frac{1}{1.5} = \frac{2}{3} \).
This mole fraction is directly used in Raoult's Law to calculate the partial pressures in a solution.

Partial Pressure

Partial pressure is the pressure exerted by a single component of a mixture of gases or vapours. It's a key concept in situations where multiple gases are mixed, such as our benzene-toluene solution.
According to Raoult's Law, the partial pressure of each component in an ideal solution is given by multiplying its mole fraction by its pure component vapour pressure.
In our example:

• The partial pressure of benzene = mole fraction of benzene \( \times \) pure benzene vapour pressure = \( \frac{2}{3} \times 75 \text{ torr} = 50 \text{ torr} \).

By understanding partial pressures, we can accurately predict how each component will contribute to the total pressure of a mixture.