Q.Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?

The key idea is Raoult's Law for ideal solutions, which states that the partial vapour pressure of each component in an ideal solution is equal to the vapour pressure of the pure component multiplied by its mole fraction in the solution.

1. First, calculate the molar masses of heptane ($C_7H_{16}$) and octane ($C_8H_{18}$). Using atomic masses C=12 g/mol and H=1 g/mol:

   $M_{\text{heptane}} = (7 \times 12) + (16 \times 1) = 84 + 16 = 100 \text{ g/mol}$

   $M_{\text{octane}} = (8 \times 12) + (18 \times 1) = 96 + 18 = 114 \text{ g/mol}$

2. Next, determine the moles of each component and their respective mole fractions:

   $n_{\text{heptane}} = \frac{26.0 \text{ g}}{100 \text{ g/mol}} = 0.260 \text{ mol}$

   $n_{\text{octane}} = \frac{35 \text{ g}}{114 \text{ g/mol}} \approx 0.3070 \text{ mol}$

   $n_{\text{total}} = 0.260 + 0.3070 = 0.5670 \text{ mol}$

   $x_{\text{heptane}} = \frac{0.260}{0.5670} \approx 0.45855$

   $x_{\text{octane}} = \frac{0.3070}{0.5670} \approx 0.54145$

3. Apply Raoult's Law to find the partial vapour pressure of each component: …

When two volatile liquids form an ideal solution, their behaviour in the vapour phase is governed by two fundamental laws: Raoult's Law and Dalton's Law of Partial Pressures.

Raoult's Law states that for an ideal solution, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction in the solution and its vapour pressure in the pure state at the same temperature. Essentially, the more of a component there is in the liquid mixture (expressed as its mole fraction), the more it contributes to the total vapour pressure above the solution.

Dalton's Law of Partial Pressures states that the total pressure exerted by a mixture of non-reacting gases (or vapours, in this case) is equal to the sum of the partial pressures of the individual gases. This means that once we find the individual contributions of heptane and octane to the vapour pressure, we simply add them up to get the total vapour pressure of the mixture.

The problem asks for the total vapour pressure of the mixture. To find this, we need to:

1. Determine the amount (in moles) of each component.
2. Calculate the mole fraction of each component in the liquid mixture.
3. Use Raoult's Law to find the partial vapour pressure of each component.
4. Sum these partial pressures using Dalton's Law to get the total vapour pressure.

Let's proceed with the calculations.

1. Identify the given information:

   • Vapour pressure of pure heptane ($P^{\circ}_{\text{heptane}}$) at 373 K = $105.2 \text{ kPa}$
   • Vapour pressure of pure octane ($P^{\circ}_{\text{octane}}$) at 373 K = $46.8 \text{ kPa}$
   • Mass of heptane ($m_{\text{heptane}}$) = $26.0 \text{ g}$
   • Mass of octane ($m_{\text{octane}}$) = $35 \text{ g}$

2. Calculate the molar masses of heptane and octane:

   Heptane is $C_7H_{16}$.

   Molar mass of heptane ($M_{\text{heptane}}$) = $(7 \times 12.011) + (16 \times 1.008) = 84.077 + 16.128 = 100.205 \text{ g/mol}$.

   Octane is $C_8H_{18}$.

   Molar mass of octane ($M_{\text{octane}}$) = $(8 \times 12.011) + (18 \times 1.008) = 96.088 + 18.144 = 114.232 \text{ g/mol}$.

3. Calculate the number of moles of each component:

   Number of moles ($n$) = Mass ($m$) / Molar mass ($M$).

   Moles of heptane ($n_{\text{heptane}}$):

Moles of octane ($n_{\text{octane}}$):

4. Calculate the mole fraction of each component in the mixture: The total number of moles ($n_{\text{total}}$) in the mixture is:

Mole fraction ($x$) of a component is its moles divided by the total moles.

Mole fraction of heptane ($x_{\text{heptane}}$):
$$x_{\text{heptane}} = \frac{n_{\text{heptane}}}{n_{\text{total}}} = \frac{0.25946 \text{ mol}}{0.56585 \text{ mol}} = 0.45859$$ …

Understanding Raoult's Law for Ideal Solutions

1. Concept First — The Idea Being Tested

This problem tests Raoult's Law for ideal solutions. The core idea is simple but powerful:

In an ideal solution, each component contributes to the total vapour pressure in proportion to its mole fraction in the liquid mixture.

Why does this happen?

Imagine a liquid surface. In a pure liquid, all molecules at the surface are of the same type, and they escape into vapour at a fixed rate (the vapour pressure). When you mix two liquids that are chemically similar (like heptane and octane — both are alkanes), the molecules at the surface are a mixture. Each type of molecule escapes at the same rate it would from its own pure liquid, but only in proportion to how many of them are at the surface — which is exactly the mole fraction.

So, for each component $i$:

where:

• $P_i$ = partial vapour pressure of component $i$ in the mixture
• $X_i$ = mole fraction of $i$ in the liquid mixture
• $P_i^\circ$ = vapour pressure of pure $i$

The total vapour pressure is simply the sum:

2. Step-by-Step Solution

Step 1: Find the molar masses

We need moles to find mole fractions. Let's calculate:

• Heptane ($C_7H_{16}$):

  $7 \times 12 + 16 \times 1 = 84 + 16 = 100$ g/mol

• Octane ($C_8H_{18}$):

  $8 \times 12 + 18 \times 1 = 96 + 18 = 114$ g/mol

Step 2: Calculate moles of each component

Given masses: 26.0 g heptane, 35.0 g octane

Step 3: Find mole fractions in the liquid

Total moles = $0.260 + 0.307 = 0.567$ mol

Check: $0.459 + 0.541 = 1.000$ ✓

Step 4: Apply Raoult's Law

Given pure vapour pressures at 373 K:

• $P_{\text{heptane}}^\circ = 105.2$ kPa
• $P_{\text{octane}}^\circ = 46.8$ kPa

Partial pressures:

Step 5: Total vapour pressure

3. Final Answer

4. Why It Works & Exam Tip

Why this works …

Here are the common mistakes students make on this exact type of problem (vapour pressure of an ideal solution of two volatile liquids), and how to avoid each.

1. Forgetting to Convert Mass to Moles First

The Mistake: Plugging the given masses (26.0 g heptane, 35 g octane) directly into Raoult's law instead of converting to moles.

How to avoid: Mass -> moles -> mole fraction -> Raoult's law. Never skip the mole conversion.

• Molar mass of heptane ($C_7H_{16}$) = $7(12) + 16(1) = 100$ g/mol -> moles = $\dfrac{26.0}{100} = 0.260$ mol
• Molar mass of octane ($C_8H_{18}$) = $8(12) + 18(1) = 114$ g/mol -> moles = $\dfrac{35}{114} \approx 0.307$ mol

2. Using Mass Fraction Instead of Mole Fraction

The Mistake: Computing $\dfrac{26}{26+35}$ and using that directly in Raoult's law.

Why it's wrong: Raoult's law depends on the mole fraction (number of molecules), not the mass fraction.

How to avoid: Always compute mole fraction from moles:

3. Using the Pure Vapour Pressure Instead of the Partial Pressure

The Mistake: Writing $P_{\text{heptane}} = 105.2$ kPa instead of $P_{\text{heptane}} = x_{\text{heptane}} \times 105.2$.

How to avoid: Always write the full form $P_i = x_i P_i^\circ$ for each component.

4. Forgetting to Add the Partial Pressures …