Q.Calculate the mass of a non-volatile solute (molar mass 40 g mol$^{-1}$) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.

The key idea is molality calculation using Raoult’s law for a non-volatile solute.

Step 1 – Relate vapour pressure lowering to mole fraction

For a non-volatile solute, Raoult’s law gives:

Here, vapour pressure is reduced to 80%, so $p = 0.80\,p^\circ$. Thus:

Step 2 – Express mole fraction in terms of masses

Molar mass of octane ($\mathrm{C_8H_{18}}$) = $8 \times 12 + 18 \times 1 = 114$ g mol$^{-1}$.

Mass of octane = 114 g, so moles of octane = $\frac{114}{114} = 1$ mol.

Let $w$ g be the mass of solute (molar mass 40 g mol$^{-1}$). Then:

Step 3 – Solve for $w$ …

1. Moles of octane ($C_8H_{18}$, $M = 114\ \text{g mol}^{-1}$):

2. Mole-fraction condition.

Raoult's law for a non-volatile solute gives $p = x_{solvent}\, p^\circ$. Since $p = 0.80\, p^\circ$,

3. Moles of solute.

4. Mass of solute ($M = 40\ \text{g mol}^{-1}$):

$w = 0.25 \times 40 = 10\ \text{g}$ …

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

This is the standard method for problems where a non-volatile solute reduces the vapour pressure of a solvent.

Step 1 — Write Raoult's Law for relative lowering

Where:

• $p^\circ$ = vapour pressure of pure solvent
• $p$ = vapour pressure of solution
• $n_2$ = moles of solute
• $n_1$ = moles of solvent

Step 2 — Interpret "reduced to 80%"

If vapour pressure reduces to 80% of $p^\circ$, then:

So the relative lowering is:

Thus:

Step 3 — Find moles of solvent (octane)

Molar mass of octane ($C_8H_{18}$) = $8 \times 12 + 18 \times 1 = 114$ g mol$^{-1}$

Given mass of octane = 114 g

--- …

Here are the common mistakes students make on this exact type of problem (mass of non-volatile solute needed to reduce vapour pressure by a given percentage), and how to avoid each.

1. Confusing "reduced to 80%" with "reduced by 80%"

The Mistake: Reading "reduce vapour pressure to 80%" as "reduce it BY 80%" (i.e. final pressure = 20% of original) instead of "final pressure = 80% of original."

How to Avoid: Read literally: "reduced to 80%" means $p = 0.80\,p^\circ$, so the relative lowering is $\dfrac{p^\circ-p}{p^\circ} = 0.20$.

2. Using the Wrong Mole-Fraction Formula

The Mistake: Writing $\dfrac{p^\circ-p}{p^\circ} = \dfrac{n_{\text{solute}}}{n_{\text{solvent}}}$ and forgetting the solute term belongs in the denominator too.

How to Avoid: The exact relation is $x_{\text{solute}} = \dfrac{n_{\text{solute}}}{n_{\text{solute}}+n_{\text{solvent}}}$; only use the dilute approximation $\approx \dfrac{n_{\text{solute}}}{n_{\text{solvent}}}$ when the problem is clearly dilute.

3. Forgetting to Convert Solvent Mass to Moles

The Mistake: Plugging 114 g of octane directly into the formula as if it were moles.

How to Avoid: Molar mass of octane ($C_8H_{18}$) = 114 g/mol, so 114 g happens to be exactly 1 mol here -- but always show the division step, don't just assume it.

4. Stopping at Moles of Solute -- Forgetting to Convert to Mass

The Mistake: Solving for $n_{\text{solute}}$ and reporting that number as the final answer instead of multiplying by the given molar mass (40 g/mol). …