1. Use Raoult’s law

For an ideal binary solution, $P_{\text{total}} = x_{\text{hexane}} P_{\text{hexane}}^0 + x_{\text{heptane}} P_{\text{heptane}}^0$ where $P^0$ denotes pure component vapour pressure.

Let

• $P_{\text{hexane}}^0 = H$
• $P_{\text{heptane}}^0 = G$

2. First solution: 1 mole hexane + 3 moles heptane

Total moles $= 4$

So mole fractions are: $x_{\text{hexane}} = \frac{1}{4}, \qquad x_{\text{heptane}} = \frac{3}{4}$

Given total vapour pressure is $550$ mm Hg: $\frac{1}{4}H + \frac{3}{4}G = 550$ Multiplying by 4: $H + 3G = 2200 \qquad (1)$

3. Second solution: add 1 more mole of heptane

Now composition becomes:

• hexane = 1 mole
• heptane = 4 moles
• total = 5 moles

New mole fractions: $x_{\text{hexane}} = \frac{1}{5}, \qquad x_{\text{heptane}} = \frac{4}{5}$

The vapour pressure increases by $10$ mm Hg, so new total pressure is: $560 \text{ mm Hg}$

Thus, $\frac{1}{5}H + \frac{4}{5}G = 560$ Multiplying by 5: $H + 4G = 2800 \qquad (2)$

4. Solve the two equations

From (2) - (1): $(H + 4G) - (H + 3G) = 2800 - 2200$ $G = 600$

So the vapour pressure of pure n-heptane is $\boxed{600 \text{ mm Hg}}$

5. Comparison with stored answer

Stored correct answer = $600$

Our derived answer matches the stored answer.