Problem 2 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

At room temperature, a dilute solution of urea is prepared by dissolving \(0.60 \mathrm{~g}\) of urea in \(360 \mathrm{~g}\) of water. If the vapour pressure of pure water at this temperature is \(35 \mathrm{mmHg}\), lowering of vapour pressure will be : (molar mass of urea \(=60 \mathrm{~g} \mathrm{~mol}^{-1}\) ) (a) \(0.027 \mathrm{mmHg}\) (b) \(0.028 \mathrm{mmHg}\) (c) \(0.017 \mathrm{mmHg}\) (d) \(0.031 \mathrm{mmHg}\)

Step-by-Step Solution

Answer

The lowering of vapor pressure is approximately 0.017 mmHg (option c).

Step 1: Calculate Moles of Urea

First, we need to calculate the number of moles of urea. We know the mass of urea, which is 0.60 g, and its molar mass is 60 g/mol. The number of moles is calculated using the formula: \[ \text{Moles of urea} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.60}{60} = 0.01 \text{ moles} \].

Step 2: Calculate Moles of Water

Next, calculate the number of moles of water. We know the mass of water, which is 360 g, and the molar mass of water is 18 g/mol. Using the formula: \[ \text{Moles of water} = \frac{\text{mass}}{\text{molar mass}} = \frac{360}{18} = 20 \text{ moles} \].

Step 3: Use Raoult's Law and Calculate Lowering of Vapor Pressure

Using Raoult's Law, the lowering of vapor pressure is calculated as:\[ \Delta P = P_0 \times X_{\text{solute}} \] where \( P_0 \) is the pure vapor pressure of water (35 mmHg), and \( X_{\text{solute}} \) is the mole fraction of the solute, urea. The mole fraction of urea is: \[ X_{\text{urea}} = \frac{\text{Moles of urea}}{\text{Moles of urea} + \text{Moles of water}} = \frac{0.01}{0.01 + 20} \approx 0.0005 \].

Step 4: Calculate the Lowering of Vapor Pressure

Substitute the values into the equation: \[ \Delta P = 35 \times 0.0005 = 0.0175 \text{ mmHg} \].

Step 5: Choose the Correct Answer

We find that the lowering of the vapor pressure is \(0.0175 \text{ mmHg} \), which is closest to option (c) 0.017 mmHg.

Key Concepts

Vapor Pressure Lowering

Vapor pressure lowering is a key concept in understanding solutions, especially when dealing with dilute solutions. It is a phenomenon observed when a non-volatile solute is added to a solvent, resulting in a decrease in the vapor pressure of the solvent. This occurs because the solute particles occupy space at the liquid's surface, reducing the number of solvent molecules that can escape into the vapor phase. This effect is explained by Raoult's Law, which states that the vapor pressure of a solvent in a solution is directly proportional to the mole fraction of the solvent.
For example, when urea is dissolved in water, it causes a lowering of the vapor pressure of water. This happens regardless of the identity of the solute as long as it is non-volatile and does not dissociate in solution. The reduction in vapor pressure can be calculated using the formula:

• \[ \Delta P = P_0 \times X_{\text{solute}} \]

where \( \Delta P \) is the change in vapor pressure, \( P_0 \) is the vapor pressure of the pure solvent, and \( X_{\text{solute}} \) is the mole fraction of the solute.

Mole Fraction

The mole fraction is a way to express the concentration of a component in a mixture and is particularly useful in the context of Raoult's Law. It is defined as the ratio of the moles of one component to the total moles of all components in the solution.
The formula for the mole fraction of a solute is:

• \[ X_{\text{solute}} = \frac{\text{moles of solute}}{\text{moles of solute} + \text{moles of solvent}} \]

In the case of urea dissolved in water, the mole fraction of urea is calculated using its moles and the moles of water.
This calculated value is used to determine how much the vapor pressure of the solvent is lowered when the solute is added. Mole fractions are dimensionless and add up to one when considering all components of the solution.

Dilute Solutions

Dilute solutions contain a small amount of solute compared to the solvent. These solutions are important when applying Raoult's Law as the assumptions simplify calculations. In dilute solutions, the solute's effect on the overall properties of the solution is minimal.
However, even small amounts of solute can significantly lower vapor pressure due to their presence at the molecular level. This is because the solute particles interfere with the escaping tendency of the solvent molecules. This characteristic makes it easy to apply formulas like Raoult's Law for finding the vapor pressure lowering in such solutions.
The example of a urea solution in water highlights how a dilute amount of solute can result in a measurable decrease in vapor pressure, which underscores the applicability of the concept to various practical and experimental laboratory scenarios.