Q.The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile, non-electrolyte solid weighing 0.5 g when added to 39.0 g of benzene (molar mass 78 g mol$^{-1}$). Vapour pressure of the solution, then, is 0.845 bar. What is the molar mass of the solid substance?

The key idea is Raoult's Law for relative lowering of vapour pressure. For a dilute solution of a non-volatile, non-electrolyte solute, the relative lowering of vapour pressure equals the mole fraction of the solute, and since $n_2 \ll n_1$ the working relation is $\frac{p^0 - p}{p^0} \approx \frac{n_2}{n_1}$ (equation 1.28) — exactly how NCERT solves this example.

1. First, calculate the moles of benzene (solvent):

   $n_1 = \frac{\text{mass of benzene}}{\text{molar mass of benzene}} = \frac{39.0 \text{ g}}{78 \text{ g mol}^{-1}} = 0.5 \text{ mol}$

2. Next, apply the approximate relation with $n_2 = \frac{w_2}{M_2} = \frac{0.5}{M_2}$:

   $\frac{p^0 - p}{p^0} = \frac{n_2}{n_1}$

   $\frac{0.850 - 0.845}{0.850} = \frac{0.5/M_2}{0.5}$

   $\frac{0.005}{0.850} = \frac{1}{M_2}$ …

When a non-volatile solute is added to a pure solvent, the vapour pressure of the resulting solution is always lower than that of the pure solvent. This phenomenon is described by Raoult's Law, which is a colligative property. Colligative properties depend only on the number of solute particles, not on their identity.

The intuition behind this is that the solute particles occupy some space at the surface of the liquid, reducing the number of solvent molecules available to escape into the vapour phase. Since fewer solvent molecules can escape, the equilibrium vapour pressure above the solution decreases. For a non-volatile solute, the solute itself does not contribute to the vapour pressure.

Raoult's Law for a solution containing a non-volatile solute states that the relative lowering of vapour pressure is equal to the mole fraction of the solute.

We can use this relationship to find the mole fraction of the unknown solid solute, and from there, its molar mass.

Here is the step-by-step solution:

1. Identify the given values and the unknown:

   • Vapour pressure of pure benzene ($P_A^0$) = $0.850 \text{ bar}$
   • Vapour pressure of the solution ($P_A$) = $0.845 \text{ bar}$
   • Mass of benzene ($w_A$) = $39.0 \text{ g}$
   • Molar mass of benzene ($M_A$) = $78 \text{ g mol}^{-1}$
   • Mass of solid solute ($w_B$) = $0.5 \text{ g}$
   • We need to find the molar mass of the solid substance ($M_B$).

2. Calculate the moles of the solvent (benzene):

   The number of moles of benzene ($n_A$) can be calculated using its given mass and molar mass.

   $n_A = \frac{w_A}{M_A}$

   $n_A = \frac{39.0 \text{ g}}{78 \text{ g mol}^{-1}}$

   $n_A = 0.5 \text{ mol}$

3. Calculate the mole fraction of the solute ($x_B$) using Raoult's Law:

   Substitute the given vapour pressures into Raoult's Law formula:

   $\frac{P_A^0 - P_A}{P_A^0} = x_B$

   $x_B = \frac{0.850 \text{ bar} - 0.845 \text{ bar}}{0.850 \text{ bar}}$

   $x_B = \frac{0.005 \text{ bar}}{0.850 \text{ bar}}$

   $x_B = \frac{5}{850}$

   $x_B = \frac{1}{170}$

4. Determine the moles of the solute ($n_B$) using the dilute-solution approximation (equation 1.28): …

Method: Raoult's Law for Relative Lowering of Vapour Pressure

This is the standard method for finding molar mass of a non-volatile solute using vapour pressure data.

Steps

Step 1: Write Raoult's Law for relative lowering

For a non-volatile solute in a volatile solvent:

where:

• $p^0 = 0.850$ bar (pure solvent vapour pressure)
• $p = 0.845$ bar (solution vapour pressure)
• $n_2$ = moles of solute (unknown)
• $n_1$ = moles of solvent

Since the solution is dilute, $n_2 \ll n_1$, so we use the approximation.

Step 2: Calculate the relative lowering

Step 3: Calculate moles of solvent (benzene)

Mass of benzene = 39.0 g

Molar mass of benzene = 78 g mol$^{-1}$

Step 4: Apply the approximation to find moles of solute

Step 5: Calculate molar mass of solute …

🧠 Concept First: What’s happening?

We have:

• Pure benzene vapour pressure: $P^\circ = 0.850 \text{ bar}$
• Solution vapour pressure: $P = 0.845 \text{ bar}$
• Solvent mass = 39.0 g, molar mass = 78 g/mol
• Solute mass = 0.5 g (non-volatile, non-electrolyte)

We use Raoult’s law for a non-volatile solute:

Since $n_2 \ll n_1$, we approximate:

Where:

• $w_2 = 0.5$ g, $w_1 = 39.0$ g, $M_1 = 78$ g/mol
• $M_2$ = molar mass of solid (to find)

✗ Mistake 1: Using molality formula directly without Raoult’s law

What students do:

They jump to $\Delta T_f = K_f \cdot m$ or similar, forgetting this is a vapour pressure problem.

Why it’s wrong:

Molality is used for boiling point elevation / freezing point depression, not directly for vapour pressure. Here, the link is Raoult’s law.

✓ How to avoid:

Always identify the physical property given (vapour pressure change) and pick the correct law. For vapour pressure lowering of a non-volatile solute, always start with:

✗ Mistake 2: Forgetting to convert masses to moles

What students do:

Plug in masses directly:

$\frac{0.850 - 0.845}{0.850} = \frac{0.5}{39.0}$ → then solve for $M_2$ incorrectly.

Why it’s wrong:

Raoult’s law uses mole ratio, not mass ratio. You must convert solvent mass to moles.

✓ How to avoid:

Always write:

Then:

✗ Mistake 3: Using the exact formula instead of the approximation

What students do:

They write:

and then try to solve a quadratic.

Why it’s not wrong but inefficient:

For dilute solutions, $n_2 \ll n_1$, so $n_1 + n_2 \approx n_1$. Using the exact form is fine but adds unnecessary algebra.

✓ How to avoid:

Check if $n_2$ is small relative to $n_1$. Here $n_1 = 0.5$ mol, and $n_2$ will be ~0.01 mol — so approximation is safe. Use:

✗ Mistake 4: Unit mismatch or forgetting to convert bar to same units

What students do:

They treat 0.850 and 0.845 as if they are in different units or forget they are already in bar.

Why it’s wrong: …