Q.An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?

Concept: Osmotic pressure is used here, but more directly this is a vapour-pressure lowering problem — the solute reduces the vapour pressure of water at its boiling point, and the observed pressure (1.004 bar) is the vapour pressure of the solution. At the normal boiling point of water (100°C), pure water exerts 1.013 bar.

Step 1 — Find the lowering of vapour pressure.

$\Delta P = P^\circ - P = 1.013 - 1.004 = 0.009 \text{ bar}$

Step 2 — Apply Raoult’s law for a dilute solution.

For a non-volatile solute,

$\frac{\Delta P}{P^\circ} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1}$ (since $n_2 \ll n_1$).

Here $n_2$ = moles of solute, $n_1$ = moles of solvent (water).

Step 3 — Express in terms of masses.

Mass of solute = 2 g per 100 g solution → solvent mass = 98 g. …

Why this works

The normal boiling point of a liquid is the temperature at which its vapour pressure equals the external atmospheric pressure (1.013 bar for water at 100 °C). When a non-volatile solute is dissolved in water, the vapour pressure of the solution is lower than that of pure water at the same temperature — this is Raoult’s law in action.

The relative lowering of vapour pressure depends only on the mole fraction of the solute, not on its chemical identity. For a dilute solution, the mole fraction of solute is approximately the ratio of moles of solute to moles of solvent. Since we know the mass percentage of the solution, we can work backwards from the pressure drop to find the molar mass of the solute.

For dilute solutions, $n_{\text{solute}} \ll n_{\text{solvent}}$, so:

Step-by-step solution

1. Identify the given data

• Solvent: water (normal boiling point = 100 °C)
• Vapour pressure of pure water at 100 °C: $P^0 = 1.013\ \text{bar}$
• Vapour pressure of solution: $P = 1.004\ \text{bar}$
• Solution is 2% by mass of non-volatile solute → 2 g solute in 100 g solution, so mass of solvent (water) = 98 g.

2. Calculate the relative lowering of vapour pressure

This dimensionless number equals the mole fraction of solute in the solution.

3. Express mole fraction in terms of moles

Let $M$ be the molar mass of the solute (in g/mol).

Moles of solute: $n_{\text{solute}} = \frac{2}{M}$

Moles of solvent (water, molar mass 18 g/mol): $n_{\text{solvent}} = \frac{98}{18} \approx 5.444\ \text{mol}$

Since the solution is dilute, $n_{\text{solute}} \ll n_{\text{solvent}}$, so:

4. Equate and solve for $M$

$\frac{2/M}{5.444} = 0.008884$ …

Method: Relative Lowering of Vapour Pressure (Raoult's Law) for Molar Mass Determination

Key Concept

At the normal boiling point of a liquid, its vapour pressure equals the external atmospheric pressure. For water, this means the vapour pressure of pure water at 100 degrees C is $1.013$ bar. When a non-volatile solute is dissolved, the vapour pressure of the solution at the same temperature (1.004 bar here) is lower -- this is Raoult's law, not osmotic pressure.

Step-by-Step Solution

Step 1: Identify given data

• Solvent: water, normal boiling point = 100 degrees C, so $P^\circ = 1.013$ bar
• Vapour pressure of the solution: $P = 1.004$ bar
• 2% solute by mass -> 2 g solute per 100 g solution -> solvent (water) mass = 98 g

Step 2: Relative lowering of vapour pressure

Step 3: Moles of solvent

Step 4: Solve for the molar mass $M$ of the solute

Let $M$ be the molar mass of the solute; $n_{\text{solute}} = \dfrac{2}{M}$. …

Here are the common mistakes students make on this exact type of problem, along with how to avoid each.

Mistake 1: Treating the given pressure as osmotic pressure

The error: The number "1.004 bar" looks like it could plug into $\pi = CRT$, so students reach for the osmotic-pressure formula.

Why it's wrong: The problem says the solution "exerts a pressure ... at the normal boiling point of the solvent." That is the vapour pressure of the solution at 100 degrees C, not an osmotic pressure -- there is no semipermeable membrane anywhere in this problem.

How to avoid:

• At the normal boiling point of a liquid, its vapour pressure equals the external (atmospheric) pressure -- for water that is $1.013$ bar.
• The solution's vapour pressure at that same temperature is 1.004 bar, lower than pure water's because of the dissolved solute. This is Raoult's law / relative lowering of vapour pressure, not osmotic pressure.

Mistake 2: Forgetting to find the vapour pressure of pure water first

The error: Students try to use 1.004 bar directly without ever bringing in the vapour pressure of pure water at 100 degrees C (1.013 bar).

How to avoid:

• Always start with: $P^\circ_{\text{water at }100^\circ\text{C}} = 1.013$ bar (this is just the external/atmospheric pressure, by the very definition of the normal boiling point).
• The lowering is: $\Delta P = P^\circ - P = 1.013 - 1.004 = 0.009$ bar.

Mistake 3: Using the mass of solution instead of the mass of solvent

The error: "2% solute" is read as "2 g solute + 100 g water," using 100 g as the solvent mass.

How to avoid:

• 2% by mass means 2 g of solute per 100 g of solution -- so the solvent (water) mass is $100 - 2 = 98$ g, not 100 g.

Mistake 4: Forgetting the dilute-solution approximation

The error: Not knowing how to relate the relative lowering of vapour pressure to mole fraction.

How to avoid: …