Q.Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea ($NH_2CONH_2$) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.

Concept: Relative Lowering of Vapour Pressure (Raoult’s Law) — the relative lowering of vapour pressure depends only on the mole fraction of the solute.

Step 1: Find moles of urea and water

Molar mass of urea ($NH_2CONH_2$) = $60 \, \text{g/mol}$.

Moles of urea = $\frac{50}{60} = 0.833 \, \text{mol}$.

Moles of water = $\frac{850}{18} = 47.22 \, \text{mol}$.

Step 2: Mole fraction of urea

$x_{\text{urea}} = \frac{0.833}{0.833 + 47.22} = \frac{0.833}{48.053} = 0.01733$.

Step 3: Relative lowering and vapour pressure

Relative lowering = $x_{\text{urea}} = 0.01733$.

Using $\frac{P^0 - P}{P^0} = x_{\text{urea}}$, we get $P = P^0 (1 - x_{\text{urea}})$.

$P = 23.8 \times (1 - 0.01733) = 23.8 \times 0.98267 = 23.39 \, \text{mm Hg}$.

Why this approach works

When a non-volatile solute like urea dissolves in water, it reduces the number of solvent molecules at the surface that can escape into vapour. Raoult’s law captures this: the vapour pressure of the solution ($p$) is directly proportional to the mole fraction of the solvent ($x_{\text{solvent}}$). The lowering of vapour pressure ($p^0 - p$) relative to the pure solvent’s vapour pressure ($p^0$) is simply the mole fraction of the solute — a neat result that avoids calculating $p$ directly if only the relative lowering is asked.

We need both the actual vapour pressure and the relative lowering, so we’ll compute mole fractions first.

Step-by-step solution

1. Find the molar masses

Urea, $NH_2CONH_2$:

$N = 14.0$, $H = 1.0$, $C = 12.0$, $O = 16.0$

Molar mass = $14.0 + 2(1.0) + 12.0 + 16.0 + 14.0 + 2(1.0) = 60.0 \text{ g/mol}$.

Water, $H_2O$:

$H = 1.0$, $O = 16.0$

Molar mass = $2(1.0) + 16.0 = 18.0 \text{ g/mol}$.

2. Calculate moles of each component

Moles of urea:

$n_{\text{urea}} = \frac{50 \text{ g}}{60.0 \text{ g/mol}} = 0.8333 \text{ mol}$.

Moles of water:

$n_{\text{water}} = \frac{850 \text{ g}}{18.0 \text{ g/mol}} = 47.222 \text{ mol}$.

3. Find mole fractions

Total moles:

$n_{\text{total}} = 0.8333 + 47.222 = 48.055 \text{ mol}$.

Mole fraction of water (solvent):

$x_{\text{water}} = \frac{47.222}{48.055} = 0.9827$.

Mole fraction of urea (solute):

$x_{\text{urea}} = \frac{0.8333}{48.055} = 0.01734$.

4. Apply Raoult’s law for vapour pressure

Raoult’s law: $p = x_{\text{water}} \cdot p^0$, where $p^0 = 23.8 \text{ mm Hg}$.

$p = 0.9827 \times 23.8 = 23.38 \text{ mm Hg}$.

Rounding to three significant figures: $23.4 \text{ mm Hg}$.

5. Compute relative lowering of vapour pressure

Relative lowering = $\frac{p^0 - p}{p^0}$.

From Raoult’s law, this equals $x_{\text{urea}}$:

$\frac{p^0 - p}{p^0} = x_{\text{urea}} = 0.01734$.

Alternatively, directly:

$\frac{23.8 - 23.38}{23.8} = \frac{0.42}{23.8} = 0.01765$ (slight difference due to rounding $p$). Using the exact mole fraction is more accurate.

6. Final values

Vapour pressure of solution: $23.4 \text{ mm Hg}$ (to three significant figures).

Relative lowering: $0.0173$ (or $1.73 \times 10^{-2}$).

Method: Raoult's Law for Non-Volatile Solute

This is a relative lowering of vapour pressure problem. The method uses Raoult's Law, which states that for a non-volatile solute (like urea), the vapour pressure of the solution depends only on the mole fraction of the solvent.

Steps

Step 1: Identify the given data

• Vapour pressure of pure water, $P^0 = 23.8 \text{ mm Hg}$
• Mass of urea (solute) = $50 \text{ g}$
• Mass of water (solvent) = $850 \text{ g}$
• Molar mass of urea ($NH_2CONH_2$) = $60 \text{ g/mol}$ (Calculation: $N=14$, $H=1$, $C=12$, $O=16$ → $2(14) + 4(1) + 12 + 16 = 60$)
• Molar mass of water ($H_2O$) = $18 \text{ g/mol}$

Step 2: Calculate moles of each component

• Moles of urea = $\frac{50}{60} = 0.833 \text{ mol}$
• Moles of water = $\frac{850}{18} = 47.22 \text{ mol}$

Step 3: Apply Raoult's Law for vapour pressure of solution

Raoult's Law:

where $\chi_{\text{solvent}}$ is the mole fraction of water.

Therefore:

Step 4: Calculate relative lowering of vapour pressure

Relative lowering is defined as:

Alternatively, this equals the mole fraction of solute:

Final Answer

Key concept: For a non-volatile solute, the relative lowering of vapour pressure depends only on the mole fraction of the solute — not on its chemical nature. This is a colligative property.

Here are the common mistakes students make when solving this molality-based vapour pressure problem, along with how to avoid each.

Mistake 1: Confusing Molality with Molarity

The error:

Students often calculate moles of solute using the volume of solvent (assuming 1 L = 1000 g) instead of using the mass of solvent in kg.

Example of the mistake:

Using 850 mL as if it were 850 g, then dividing by 1000 to get 0.85 L — but molality needs kg of solvent, not litres.

How to avoid:

• Molality ($m$) = $\frac{\text{moles of solute}}{\text{mass of solvent in kg}}$
• Here, solvent mass = 850 g = 0.850 kg (not 0.850 L).
• Always convert grams of solvent to kilograms by dividing by 1000.

Mistake 2: Incorrect Molar Mass of Urea

The error:

Using wrong atomic masses or forgetting that urea has two nitrogen atoms.

Correct molar mass:

Urea = $NH_2CONH_2$

• N: 2 × 14 = 28
• H: 4 × 1 = 4
• C: 1 × 12 = 12
• O: 1 × 16 = 16 Total = 60 g/mol

How to avoid:

• Write the molecular formula clearly.
• Double-check each element’s count.
• Use standard atomic masses (N=14, H=1, C=12, O=16).

Mistake 3: Forgetting that Urea is Non-Volatile and Non-Electrolyte

The error:

Assuming urea dissociates or contributes to vapour pressure directly.

Why it matters:

• Urea does not dissociate in water (no ions).
• It is non-volatile, so only water contributes to vapour pressure.
• Raoult’s law applies: $P_{\text{solution}} = X_{\text{water}} \cdot P_{\text{water}}^0$

How to avoid:

• Remember: For non-volatile solutes, vapour pressure lowering depends only on mole fraction of solvent.
• No van’t Hoff factor ($i=1$) needed here.

Mistake 4: Using Mass of Solution Instead of Solvent in Mole Fraction

The error:

Calculating mole fraction using total mass of solution instead of moles of each component.

Correct approach:

• Moles of water = $\frac{850}{18} = 47.22$ mol
• Moles of urea = $\frac{50}{60} = 0.833$ mol
• Mole fraction of water = $\frac{47.22}{47.22 + 0.833}$

How to avoid:

• Always convert masses to moles first.
• Mole fraction = $\frac{\text{moles of component}}{\text{total moles}}$.

Mistake 5: Confusing Relative Lowering with Absolute Lowering

The error:

Stating relative lowering as $P^0 - P$ (which is absolute lowering) instead of $\frac{P^0 - P}{P^0}$.

Correct definitions:

• Absolute lowering = $P^0 - P$
• Relative lowering = $\frac{P^0 - P}{P^0} = X_{\text{solute}}$

How to avoid:

• Memorise: Relative lowering = mole fraction of solute.
• For this problem: $X_{\text{urea}} = \frac{0.833}{47.22 + 0.833} \approx 0.0173$ So relative lowering = 0.0173 (dimensionless).

Mistake 6: Forgetting Units in Final Answer

The error:

Giving vapour pressure without units, or using wrong units (e.g., atm instead of mm Hg).

How to avoid:

• The problem gives $P^0 = 23.8$ mm Hg — answer must be in mm Hg.
• Final vapour pressure: $P = P^0 \times X_{\text{water}} = 23.8 \times \frac{47.22}{48.053} \approx 23.8 \times 0.9827 \approx 23.4$ mm Hg

Quick Checklist to Avoid All Mistakes

Final Answer (for reference):

• Vapour pressure of solution ≈ 23.4 mm Hg
• Relative lowering ≈ 0.0173 (or 1.73%)