At 300 K, the vapour pressure of a solution containing 1 mole of n-hexane and 3 moles of n-heptane i

Q: Why do we use mole fractions instead of just moles?

Raoult's law relates partial pressure to mole fraction because the mole fraction represents the probability of a specific molecule being at the surface and escaping into the vapour phase.

Q: Why did the total pressure increase when we added more heptane?

Because we added more of a volatile component, increasing its mole fraction and thus its contribution to the total vapour pressure of the solution.

Q: Can we solve this without multiplying by 0.20 and 0.25?

Absolutely! You can use the substitution method by isolating $P^\circ_{\text{hex}}$ in the first equation and plugging it into the second. However, the elimination method is often faster and less prone to algebraic errors.

Q: What exactly does $P^\circ$ represent?

It represents the vapour pressure of the pure liquid at that specific temperature, meaning the pressure it would exert if no other substances were mixed with it.

Q: Is this an ideal solution?

Yes, Raoult's law strictly applies to ideal solutions where the intermolecular forces between different molecules (hexane-heptane) are similar to those between identical molecules (hexane-hexane or heptane-heptane).

Question

At 300 K, the vapour pressure of a solution containing 1 mole of n-hexane and 3 moles of n-heptane is 550 mm of Hg. At the same temperature, if one more mole of n-heptane is added to this solution, the vapour pressure of the solution increases by 10 mm of Hg. What is the vapour pressure in mm Hg of n-heptane in its pure state ........... ?

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a laboratory with two closed containers. In the first container, we have a mixture of $1$ mole of n-hexane and $3$ moles of n-heptane. The total vapour pressure inside this container is measured to be $550$ mm of Hg.

Now, we perform a simple operation: we add exactly $1$ more mole of n-heptane to the mixture, bringing the total moles of heptane to $4$. As a result, the total vapour pressure increases by $10$ mm of Hg, making the new pressure $560$ mm of Hg. Our goal is to find the vapour pressure of pure n-heptane at this temperature.

### The Master Equation: Raoult's Law

To solve this, we need to invoke **Raoult's Law**. It states that the total vapour pressure of an ideal solution is the sum of the partial pressures of its volatile components. Mathematically, the partial pressure of a component is its pure vapour pressure multiplied by its mole fraction in the solution.

$$P_{	ext{total}} = P^\circ_{	ext{hex}} \chi_{	ext{hex}} + P^\circ_{	ext{hep}} \chi_{	ext{hep}}$$

Here, $P^\circ$ represents the pure vapour pressure, and $\chi$ represents the mole fraction.

### Setting Up the Equations

Let's apply Raoult's Law to our two scenarios.

**Case I:**
The total number of moles is $1 + 3 = 4$.
The mole fraction of hexane is $\chi_{	ext{hex}} = \frac{1}{4} = 0.25$.
The mole fraction of heptane is $\chi_{	ext{hep}} = \frac{3}{4} = 0.75$.
Substituting these into our master equation, we get:
$$550 = 0.25 P^\circ_{	ext{hex}} + 0.75 P^\circ_{	ext{hep}}$$

**Case II:**
After adding $1$ mole of heptane, the total number of moles becomes $1 + 4 = 5$.
The new mole fraction of hexane is $\chi_{	ext{hex}} = \frac{1}{5} = 0.20$.
The new mole fraction of heptane is $\chi_{	ext{hep}} = \frac{4}{5} = 0.80$.
The new total pressure is $560$ mm of Hg. Substituting these values, we get our second equation:
$$560 = 0.20 P^\circ_{	ext{hex}} + 0.80 P^\circ_{	ext{hep}}$$

### Solving the System

We now have a system of two linear equations with two variables ($P^\circ_{	ext{hex}}$ and $P^\circ_{	ext{hep}}$). To find $P^\circ_{	ext{hep}}$, we can eliminate $P^\circ_{	ext{hex}}$.

Let's multiply the first equation by $0.20$ and the second equation by $0.25$ to equalize the coefficients of $P^\circ_{	ext{hex}}$.

Equation 1 $	imes 0.20$:
$$110 = 0.05 P^\circ_{	ext{hex}} + 0.15 P^\circ_{	ext{hep}}$$

Equation 2 $	imes 0.25$:
$$140 = 0.05 P^\circ_{	ext{hex}} + 0.20 P^\circ_{	ext{hep}}$$

### Final Calculation

Now, we simply subtract the modified first equation from the modified second equation. The $P^\circ_{	ext{hex}}$ terms cancel out perfectly!

$$140 - 110 = (0.05 - 0.05)P^\circ_{	ext{hex}} + (0.20 - 0.15)P^\circ_{	ext{hep}}$$
$$30 = 0.05 P^\circ_{	ext{hep}}$$

Finally, dividing both sides by $0.05$:
$$P^\circ_{	ext{hep}} = \frac{30}{0.05} = 600 	ext{ mm of Hg}$$

And there we have it! The vapour pressure of pure n-heptane is **$600$ mm of Hg**. The elegance of linear algebra combined with physical chemistry principles leads us straight to the answer.

Enter Numerical Value:

Visualized Solution

Visualizing the Two States

Raoult's Law

Setting up Case I

Setting up Case II

Equating Coefficients

Eliminating $P_{hex}^{\circ}$

Calculating $P_{hep}^{\circ}$

Finding $P_{hex}^{\circ}$

The Sigma Insight: Henry's Law and Raoult's Law

Analyzing the Setup

The Master Equation

Setting Up the Equations

Solving the System

Final Calculation

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