Question

The relative lowering of vapour pressure produced by dissolving 18 g of urea (Molar mass = 60 g mol\(^{-1}\)) in 100 g of water is

• A. 0.025
• B. 0.5
• C. 0.05
• D. 0.25

Hint:

For dilute solutions, the mole fraction of the solute can be approximated as \(x_{solute} \approx \frac{n_{solute}}{n_{solvent}}\). Let's check this approximation here: \( \frac{0.3}{5.55} \approx 0.054 \). This is also close to 0.05. This approximation is useful for quick calculations in multiple-choice questions.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
According to Raoult's Law, the Relative Lowering of Vapour Pressure (RLVP) of a solution containing a non-volatile solute is equal to the mole fraction of the solute.
Step 2: Key Formula or Approach:
Use the formula: $RLVP = \frac{P^{0} - P}{P^{0}} = X_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}}$.
Calculate moles of urea and moles of water first.
Step 3: Detailed Explanation:
1. Calculate Moles of Urea ($n_{\text{urea}}$):
Mass given = 18 g, Molar Mass = 60 g/mol.
\[ n_{\text{urea}} = \frac{18 \text{ g}}{60 \text{ g/mol}} = 0.3 \text{ mol}. \]

2. Calculate Moles of Water ($n_{\text{water}}$):
Mass given = 100 g, Molar Mass = 18 g/mol.
\[ n_{\text{water}} = \frac{100 \text{ g}}{18 \text{ g/mol}} \approx 5.55 \text{ mol}. \]

3. Calculate Mole fraction of Urea ($X_{\text{urea}}$):
\[ X_{\text{urea}} = \frac{n_{\text{urea}}}{n_{\text{urea}} + n_{\text{water}}} \]
\[ X_{\text{urea}} = \frac{0.3}{0.3 + 5.55} = \frac{0.3}{5.85} \]
\[ X_{\text{urea}} \approx 0.051 \approx 0.0
5. \]
Step 4: Final Answer:
The relative lowering of vapour pressure is approximately 0.0
5.