Problem 55 - A Complete Resource Book in Chemistry for JEE Main

Question

At room temperature, the mole fraction of a solute is \(0.25\) and the vapour pressure of a solvent is \(0.80 \mathrm{~atm}\). The lowering of vapour pressure is (a) \(0.60\) (b) \(0.75\) (c) \(0.20\) (d) \(0.80\).

Step-by-Step Solution

Answer

The lowering of vapour pressure is (c) 0.20.

Step 1: Understanding the Problem

We are given the mole fraction of a solute as 0.25 and the vapor pressure of a pure solvent as 0.80 atm. We need to find the lowering of vapour pressure due to the solute.

Step 2: Applying Raoult's Law

Raoult's Law states that the vapor pressure of a solvent in a solution ( P_{ ext{solution}} ) is the product of the mole fraction of the solvent ( x_{ ext{solvent}} ) and the vapor pressure of the pure solvent ( P_{ ext{solvent}} ). It can be formulated as:\[ P_{ ext{solution}} = x_{ ext{solvent}} imes P_{ ext{solvent}}\]

Step 3: Calculate the Mole Fraction of the Solvent

The mole fraction of the solvent ( x_{ ext{solvent}} ) can be calculated by subtracting the mole fraction of the solute from 1:\[ x_{ ext{solvent}} = 1 - x_{ ext{solute}} = 1 - 0.25 = 0.75\]

Step 4: Calculate the Vapor Pressure of the Solution

Using Raoult's Law, we calculate the vapor pressure of the solution:\[ P_{ ext{solution}} = x_{ ext{solvent}} imes P_{ ext{solvent}} = 0.75 imes 0.80 = 0.60 ext{ atm}\]

Step 5: Determine the Lowering of Vapor Pressure

The lowering of vapor pressure is given by the difference between the vapor pressure of the pure solvent and the vapor pressure of the solution:\[ ext{Lowering of Vapor Pressure} = P_{ ext{solvent}} - P_{ ext{solution}} = 0.80 - 0.60 = 0.20 ext{ atm}\]

Key Concepts

Vapour Pressure of Solutions

Understanding vapour pressure in solutions is key to grasping Raoult's Law. Vapour pressure is the pressure exerted by a vapor in equilibrium with its liquid or solid form in a closed system. It's an indicator of a substance's tendency to evaporate. When a solute is added to a solvent, the solution's vapour pressure is generally lower than that of the pure solvent. This phenomenon occurs because the solute particles interfere with the evaporation process, resulting in fewer solvent molecules escaping into the vapor phase. Consequently, the overall pressure exerted by the vapor above the solution decreases. This effect plays a vital role in various practical and industrial processes.

Mole Fraction

The mole fraction is a way of expressing the concentration of a component in a mixture. It's defined as the ratio of the number of moles of a component to the total number of moles in the mixture. Calculating mole fractions in a solution involves two components: the solute and the solvent. Sum of their mole fractions always equals one.

• For the solvent, the mole fraction is given by:
  \( x_{\text{solvent}} = \frac{\text{moles of solvent}}{\text{total moles in solution}} \).
• For the solute, it's:
  \( x_{\text{solute}} = \frac{\text{moles of solute}}{\text{total moles in solution}} \).

In the context of Raoult’s Law, the mole fraction of the solvent is crucial for determining the vapour pressure of a solution. By knowing this ratio, we can predict how the solution behaves relative to the pure solvent.

Lowering of Vapour Pressure

Lowering of vapour pressure is a direct result of the addition of a solute to a solvent. According to Raoult's Law, when a non-volatile solute is dissolved in a volatile solvent, the vapor pressure of the solvent above the solution is decreased. Here's how it works:

• Initially, the pure solvent has a certain vapour pressure.
• When you add a solute, it reduces the solvent's mole fraction in the solution.
• The vapor pressure of the solution is then calculated by multiplying this reduced mole fraction by the pure solvent's vapor pressure.
• The decrease in vapour pressure, or "lowering of vapour pressure," is the difference between these two pressures.

This concept is utilized in various applications, such as antifreeze solutions in cooling systems or salting roads to prevent ice formation. Recognizing how solute concentration affects pressure is essential for these real-world uses.