Q.The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.

Concept: Raoult's law for dilute solutions relates vapour pressure lowering to mole fraction of solute.

For a non-volatile solute, the vapour pressure of solution is given by:

where $P_0 = 12.3 \text{ kPa}$ is the vapour pressure of pure water.

A 1 molal solution contains 1 mol solute in 1000 g water. Moles of water = $\frac{1000}{18} = 55.56$ mol.

Mole fraction of water: …

When a non-volatile solute dissolves in a solvent, it occupies space at the surface and reduces the number of solvent molecules that can escape into the vapour phase. This phenomenon—vapour pressure lowering—is quantified by Raoult's law, which states that the vapour pressure of the solution is proportional to the mole fraction of the solvent.

The key insight is that molality tells us moles of solute per kilogram of solvent, so we can convert this to mole fractions and then apply Raoult's law directly.

where $\chi_{\text{solvent}}$ is the mole fraction of the solvent and $P^\circ_{\text{solvent}}$ is the vapour pressure of the pure solvent.

Step-by-step calculation

1. Identify what 1 molal means.

   A 1 molal solution contains 1 mole of solute dissolved in 1 kg (1000 g) of water. We need to find how many moles of water that corresponds to.

2. Calculate moles of water.

   The molar mass of water is 18 g/mol, so:

3. Find the mole fraction of water. The total moles in solution = moles of water + moles of solute = $55.56 + 1 = 56.56$ mol.

4. Apply Raoult's law. …

Vapour Pressure of a 1 Molal Solution – Raoult’s Law Application

1. Concept First – The Idea Being Tested

This question tests Raoult’s Law for a non-volatile solute. The core idea is simple:

When you dissolve a non-volatile solute (one that doesn’t evaporate) in a solvent, the solute particles block some solvent molecules from escaping into the vapour phase. This lowers the vapour pressure of the solvent.

Why does this happen?

• In pure water, all surface molecules can evaporate freely.
• When solute is added, some surface sites are occupied by solute particles (which don’t evaporate).
• Fewer water molecules are available at the surface to escape → vapour pressure drops.

Raoult’s Law quantifies this:

Where:

• $P_{\text{solution}}$ = vapour pressure of the solution
• $X_{\text{solvent}}$ = mole fraction of the solvent
• $P_{\text{solvent}}^0$ = vapour pressure of pure solvent (given as 12.3 kPa)

The key challenge here: we are given molality (1 molal), not mole fraction. So we must convert.

2. Step-by-Step Solution

Step 1: Understand what “1 molal” means

A 1 molal solution means:

• 1 mole of solute dissolved in 1 kg of solvent (water).

So we have:

• Moles of solute, $n_{\text{solute}} = 1 \text{ mol}$
• Mass of water = 1000 g

Step 2: Find moles of water (solvent)

Molar mass of water ($H_2O$) = 18 g/mol

Step 3: Calculate mole fraction of water

Mole fraction of solvent (water) is:

Substitute:

Calculate:

Why this step?

Raoult’s Law uses mole fraction, not molality. We must convert because the “blocking effect” depends on the relative number of solute vs solvent particles.

Step 4: Apply Raoult’s Law

Given $P_{\text{water}}^0 = 12.3 \text{ kPa}$:

3. Final Answer

The vapour pressure of the 1 molal solution is $\boxed{12.08 \text{ kPa}}$.

4. Why It Works & Exam Tip

Why it works …

Here are the common mistakes students make on this exact type of problem (finding the vapour pressure of a solution given its molality), and how to avoid each.

1. Using Molality Directly in Raoult's Law

The Mistake: Plugging molality straight into $P = x_{\text{solvent}} \cdot P^\circ$, forgetting that Raoult's law needs mole fraction, not molality.

How to avoid: Always convert molality -> moles of solvent -> mole fraction first.

2. Forgetting What "1 Molal" Means

The Mistake: Treating "1 molal solute" as "1 mole of solute in the whole solution" rather than "1 mole of solute per kilogram of solvent."

How to avoid: 1 molal = 1 mol solute dissolved in exactly 1000 g (1 kg) of solvent (water here).

3. Using the Mole Fraction of Solute Instead of Solvent

The Mistake: Writing $P = x_{\text{solute}} \cdot P^\circ$ by mistake.

How to avoid: Raoult's law for the vapour pressure of the solution uses the mole fraction of the solvent:

4. Forgetting That Water's Molar Mass Gives About 55.5 mol per kg

The Mistake: Miscalculating moles of water in 1000 g, e.g. forgetting to divide by 18. …