Benzene and toluene form nearly ideal solutions. At 20°C, the vapour pressure of benzene is 75 torr

Q: Why do we ignore the vapour pressure of toluene when calculating the partial pressure of benzene?

According to Dalton's Law and Raoult's Law, the partial pressure of a specific component in an ideal solution depends only on its own pure vapour pressure and its own mole fraction. Toluene's vapour pressure contributes to the *total* pressure, but not to benzene's *partial* pressure.

Q: How do I quickly remember the molar masses of benzene and toluene?

Benzene is $C_6H_6$, so $12 \times 6 + 6 = 78\text{ g/mol}$. Toluene is simply benzene with one hydrogen replaced by a methyl group ($-CH_3$), making it $C_6H_5CH_3$. Its molar mass is $78 - 1 + 15 = 92\text{ g/mol}$.

Q: What does 'nearly ideal solution' mean here?

An ideal solution is one that perfectly obeys Raoult's Law across all concentrations. It implies that the intermolecular forces between benzene-benzene, toluene-toluene, and benzene-toluene are all roughly equal.

Q: Why is the temperature ($20^\circ\text{C}$) given if we don't use it in the formula?

Vapour pressure is highly temperature-dependent. The values $75\text{ torr}$ and $22\text{ torr}$ are specifically valid only at $20^\circ\text{C}$. It's provided to ensure scientific accuracy, even though it doesn't enter the algebraic calculation.

Q: Could I have solved this by finding the total pressure first?

You could find the total pressure ($p_{total} = p_b + p_t$), but it's an unnecessary extra step since the question only asks for the partial pressure of benzene. Always aim for the most direct path to save time in exams!

Question

Benzene and toluene form nearly ideal solutions. At $20^\circ	ext{C}$, the vapour pressure of benzene is $75	ext{ torr}$ and that of toluene is $22	ext{ torr}$. The partial vapour pressure of benzene at $20^\circ	ext{C}$ for a solution containing $78	ext{ g}$ of benzene and $46	ext{ g}$ of toluene in torr is

Options:

(A) 53.5

(B) 37.5

(C) 25

(D) 50

Sigma Percentile · Accepted Answer

### The Magic of Ideal Solutions

Imagine you are standing in a laboratory, looking at a beaker containing a perfectly mixed blend of benzene and toluene. These two liquids are like best friends; their molecules are so similar in size and structure that they mix seamlessly, forming what we call a **nearly ideal solution**.

In an ideal solution, the intermolecular forces between the different molecules (benzene-toluene) are almost identical to the forces between the same molecules (benzene-benzene or toluene-toluene). Because of this harmony, they perfectly obey **Raoult's Law**.

### The Master Equation: Raoult's Law

Raoult's Law is the golden key to unlocking this problem. It states that the partial vapour pressure of any volatile component in an ideal solution is equal to the product of its pure vapour pressure and its mole fraction in the solution.

Mathematically, for benzene, this is written as:
$$p_b = p_b^\circ \chi_b$$

We are already given the pure vapour pressure of benzene, $p_b^\circ = 75	ext{ torr}$. Our only mission now is to find its mole fraction, $\chi_b$.

### Counting the Molecules

To find the mole fraction, we first need to know exactly how many moles of each substance are swimming in our beaker. Let's calculate them one by one.

For benzene ($C_6H_6$), the molar mass is $12 	imes 6 + 1 	imes 6 = 78	ext{ g/mol}$.
$$n_b = \frac{	ext{Mass}}{	ext{Molar Mass}} = \frac{78	ext{ g}}{78	ext{ g/mol}} = 1	ext{ mol}$$

For toluene ($C_6H_5CH_3$), the molar mass is $12 	imes 7 + 1 	imes 8 = 92	ext{ g/mol}$.
$$n_t = \frac{	ext{Mass}}{	ext{Molar Mass}} = \frac{46	ext{ g}}{92	ext{ g/mol}} = 0.5	ext{ mol}$$

### The Final Calculation

Now that we have the moles, finding the mole fraction of benzene is a breeze. It is simply the moles of benzene divided by the total moles in the mixture.

$$\chi_b = \frac{n_b}{n_b + n_t} = \frac{1}{1 + 0.5} = \frac{1}{1.5} = \frac{2}{3}$$

With the mole fraction in hand, we return to our master equation, Raoult's Law, to find the final partial pressure.

$$p_b = 75 	imes \frac{2}{3} = 50	ext{ torr}$$

And there we have it! The partial vapour pressure of benzene in this mixture is exactly $50	ext{ torr}$. Notice how elegant and straightforward the math becomes when you understand the physical reality behind the formulas.

Select Answer:

Visualized Solution

$p_{b}^{\circ} = 75 torr, p_{t}^{\circ} = 22 torr$

$p_{b} = p_{b}^{\circ} χ_{b}$

$n_{b} = \frac{w _{b}}{M _{b}}$

$n_{t} = \frac{w _{t}}{M _{t}}$

$χ_{b} = \frac{n _{b}}{n _{b} + n _{t}}$

$p_{b} = 75 \times \frac{2}{3}$

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

The Magic of Ideal Solutions

The Master Equation

Counting the Molecules

The Final Calculation

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