On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liqu

Q: Why do we assume the solution is ideal?

The problem explicitly states it is an ideal solution. Physically, heptane and octane are structurally similar non-polar hydrocarbons, meaning their intermolecular forces are nearly identical, which is the primary condition for forming an ideal solution.

Q: What if the masses were given in different units?

You must always convert mass to grams to use the standard molar mass in g/mol for calculating moles. Moles are the fundamental unit required for mole fractions.

Q: Can I use mass fractions instead of mole fractions in Raoult's Law?

No, Raoult's Law strictly depends on the number of particles at the surface of the liquid, which is represented by mole fractions, not mass.

Q: Why is the total pressure between the two pure vapor pressures?

The total pressure of an ideal binary solution is a weighted average of the pure vapor pressures based on their mole fractions, so mathematically, it must always lie between the two pure values.

Q: How would this change if the solution showed positive deviation?

If the solution showed positive deviation, the actual total vapor pressure would be higher than the value calculated using Raoult's Law due to weaker intermolecular forces in the mixture compared to the pure liquids.

Question

On mixing, heptane and octane form an ideal solution. At $373	ext{ K}$, the vapour pressures of the two liquid components (heptane and octane) are $105	ext{ kPa}$ and $45	ext{ kPa}$, respectively. Vapour pressure of the solution obtained by mixing $25	ext{ g}$ of heptane and $35	ext{ g}$ of octane will be (molar mass of heptane $= 100	ext{ g mol}^{-1}$ and of octane $= 114	ext{ g mol}^{-1}$).

Options:

(A) $72.0	ext{ kPa}$

(B) $36.1	ext{ kPa}$

(C) $96.2	ext{ kPa}$

(D) $144.5	ext{ kPa}$

Sigma Percentile · Accepted Answer

Have you ever wondered what happens when you mix two volatile liquids together? Do they fight for space in the vapor phase, or do they cooperate? In this thrilling problem, we are going to mix heptane and octane—two classic hydrocarbons—and predict the exact pressure they exert together.

### Analyzing the Setup

Imagine a closed beaker. Inside, we pour $25 	ext{ g}$ of heptane and $35 	ext{ g}$ of octane. Because these two molecules are structurally very similar (both are straight-chain alkanes), they interact with each other almost exactly as they interact with themselves. This means they form an **ideal solution**.

For an ideal solution, there are no unexpected energy changes or volume changes upon mixing. The pure vapor pressure of heptane at $373 	ext{ K}$ is given as $p_H^\circ = 105 	ext{ kPa}$, and for octane, it is $p_O^\circ = 45 	ext{ kPa}$. Our mission is to find the total vapor pressure, $p_T$, of this newly formed mixture.

### The Master Equation: Raoult's Law

To solve this, we need a powerful tool: **Raoult's Law**. This law states that the partial vapor pressure of any volatile component in an ideal solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution.

Mathematically, the total pressure is the sum of the partial pressures:

$$p_T = p_H + p_O$$

Substituting Raoult's Law for each component, we get our master equation:

$$p_T = X_H p_H^\circ + X_O p_O^\circ$$

Here, $X_H$ and $X_O$ are the mole fractions of heptane and octane, respectively. We already know the pure vapor pressures, so our next immediate goal is to find these mole fractions.

### The Currency of Chemistry: Moles

Before we can find mole fractions, we must convert our given masses into moles. Moles are the true currency of chemistry because they tell us the actual number of particles we are dealing with.

For heptane, the molar mass is $M_H = 100 	ext{ g mol}^{-1}$. Let's calculate its moles:

$$n_H = \frac{w_H}{M_H} = \frac{25}{100} = 0.25 	ext{ mol}$$

For octane, the molar mass is $M_O = 114 	ext{ g mol}^{-1}$. Let's calculate its moles:

$$n_O = \frac{w_O}{M_O} = \frac{35}{114} \approx 0.307 	ext{ mol}$$

### Finding the Mole Fractions

Now that we have the moles, finding the mole fraction is straightforward. The mole fraction of heptane, $X_H$, is the ratio of the moles of heptane to the total moles in the solution.

$$X_H = \frac{n_H}{n_H + n_O}$$

Substituting our values:

$$X_H = \frac{0.25}{0.25 + 0.307} = \frac{0.25}{0.557} \approx 0.45$$

Since the solution only contains these two components, the sum of their mole fractions must be exactly $1$. Therefore, the mole fraction of octane is simply:

$$X_O = 1 - X_H = 1 - 0.45 = 0.55$$

### The Final Calculation

We have successfully gathered all the pieces of the puzzle. It is time to substitute them back into our master equation.

$$p_T = (0.45)(105) + (0.55)(45)$$

Let's break down the math. The partial pressure of heptane is $0.45 	imes 105 = 47.25 	ext{ kPa}$. The partial pressure of octane is $0.55 	imes 45 = 24.75 	ext{ kPa}$.

Adding these together gives us the final total vapor pressure:

$$p_T = 47.25 + 24.75 = 72.0 	ext{ kPa}$$

**The final total vapor pressure of the solution is $72.0 	ext{ kPa}$.** Notice how this value lies perfectly between the pure vapor pressures of the two components ($45 	ext{ kPa}$ and $105 	ext{ kPa}$). This is a hallmark of an ideal binary solution!

Select Answer:

Visualized Solution

$Ideal Solution Setup$

$Raoult’s Law$

$Calculating Moles$

$Moles Values$

$Mole Fractions$

$Substituting in Raoult’s Law$

$Total Vapor Pressure$

$What’s Next?$

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

Analyzing the Setup

The Master Equation

The Currency of Chemistry

Finding the Mole Fractions

The Final Calculation

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