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: Key Formula for Relative Lowering of Vapour Pressure (RLVP).
According to Raoult's law, the relative lowering of vapour pressure of a solution containing a non-volatile solute is equal to the mole fraction of the solute (\(x_{solute}\)). \[ \text{RLVP} = \frac{P^0 - P}{P^0} = x_{solute} \] Where: \[ x_{solute} = \frac{\text{moles of solute}}{\text{moles of solute} + \text{moles of solvent}} \] Step 2: Calculate the moles of solute (urea) and solvent (water).
Moles of urea (n\(_1\)):

• Mass of urea = 18 g
• Molar mass of urea = 60 g/mol

\[ n_1 = \frac{\text{mass}}{\text{molar mass}} = \frac{18 \text{ g}}{60 \text{ g/mol}} = 0.3 \text{ mol} \] Moles of water (n\(_2\)):

• Mass of water = 100 g
• Molar mass of water (H\(_2\)O) = 18 g/mol

\[ n_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{100 \text{ g}}{18 \text{ g/mol}} \approx 5.55 \text{ mol} \] Step 3: Calculate the mole fraction of the solute (urea).
\[ x_{urea} = \frac{n_1}{n_1 + n_2} = \frac{0.3}{0.3 + 5.55} = \frac{0.3}{5.85} \] \[ x_{urea} \approx 0.05128 \] Step 4: Final Answer.
The relative lowering of vapour pressure is equal to the mole fraction of the solute, which is approximately 0.05.