Liquids A and B form an ideal solution in the entire composition range. At 350 K, the vapour pressur

Q: Why does the question use $x_A$ and $x_B$ for the vapour phase in the options?

Standard convention uses $x$ for the liquid phase and $y$ for the vapour phase. However, sometimes questions use $x$ generically for mole fraction in any phase. Always read the context carefully!

Q: What does "ideal solution" mean in this context?

An ideal solution strictly obeys Raoult's Law across all concentrations. It implies that the intermolecular forces between A-A, B-B, and A-B are identical.

Q: How do I know which component is more volatile?

The component with the higher pure vapour pressure ($p^\circ$) at a given temperature is more volatile. Here, B is more volatile than A.

Q: Can I solve this without calculating the total pressure?

You can find the ratio $y_A / y_B = p_A / p_B$, which gives $2.8 / 7.2 = 28 / 72 = 7 / 18$. Since $y_A + y_B = 1$, you can solve for $y_A$ and $y_B$ from this ratio, but calculating total pressure is usually more straightforward.

Q: What if the solution was non-ideal?

If it were non-ideal, Raoult's Law wouldn't hold directly. You would need to account for positive or negative deviations, often using activity coefficients, which is beyond the scope of this specific problem.

Question

Liquids $A$ and $B$ form an ideal solution in the entire composition range. At $350	ext{ K}$, the vapour pressures of pure $A$ and pure $B$ are $7 	imes 10^3	ext{ Pa}$ and $12 	imes 10^3	ext{ Pa}$, respectively. The composition of the vapour in equilibrium with a solution containing $40$ mole percent of $A$ at this temperature is

Options:

(A) $x_A = 0.76; x_B = 0.24$

(B) $x_A = 0.28; x_B = 0.72$

(C) $x_A = 0.4; x_B = 0.6$

(D) $x_A = 0.37; x_B = 0.63$

Sigma Percentile · Accepted Answer

The problem of finding the composition of a vapour phase in equilibrium with an ideal liquid solution is a classic application of two fundamental laws of physical chemistry: **Raoult's Law** and **Dalton's Law**.

Let's break down the journey of solving this problem step-by-step.

### Analyzing the Setup

Imagine a closed container where a liquid mixture of two volatile components, $A$ and $B$, is in perfect equilibrium with its vapour. We are given the composition of the liquid phase. The problem states that the solution contains $40$ mole percent of $A$.

This directly gives us the mole fraction of $A$ in the liquid phase:
$x_A = 0.4$

Since it's a binary mixture, the mole fraction of $B$ must make up the rest:
$x_B = 1 - 0.4 = 0.6$

We are also given the pure vapour pressures of both components at the given temperature ($350	ext{ K}$):
$p_A^\circ = 7 	imes 10^3	ext{ Pa}$
$p_B^\circ = 12 	imes 10^3	ext{ Pa}$

### The Master Equation: Raoult's Law

To connect the liquid composition to the vapour pressure, we use **Raoult's Law**. It states that the partial vapour pressure of each component in an ideal solution is the product of its pure vapour pressure and its mole fraction in the liquid phase.

Let's calculate the partial pressure for component $A$:
$p_A = x_A p_A^\circ = 0.4 	imes 7 	imes 10^3 = 2.8 	imes 10^3	ext{ Pa}$

Now, let's do the same for component $B$:
$p_B = x_B p_B^\circ = 0.6 	imes 12 	imes 10^3 = 7.2 	imes 10^3	ext{ Pa}$

The total pressure of the vapour mixture is simply the sum of these partial pressures:
$p_{	ext{total}} = p_A + p_B = (2.8 + 7.2) 	imes 10^3 = 10.0 	imes 10^3	ext{ Pa}$

### Dalton's Law: The Vapour's Perspective

We have the pressures, but the question asks for the composition of the vapour phase. This is where **Dalton's Law of Partial Pressures** steps in. According to Dalton's Law, the mole fraction of a gas in a mixture (let's call it $y$) is its partial pressure divided by the total pressure.

For component $A$, the mole fraction in the vapour phase is:
$y_A = \frac{p_A}{p_{	ext{total}}} = \frac{2.8 	imes 10^3}{10.0 	imes 10^3} = 0.28$

For component $B$, the mole fraction in the vapour phase is:
$y_B = \frac{p_B}{p_{	ext{total}}} = \frac{7.2 	imes 10^3}{10.0 	imes 10^3} = 0.72$

*(Note: The options in the question use the variable $x$ to denote the vapour phase composition, which is a slight deviation from standard notation, but the numerical values $0.28$ and $0.72$ perfectly match option (b).)*

### The Volatility Rule

Before we wrap up, look closely at the results. The liquid had $60\%$ of component $B$ ($x_B = 0.6$), but the vapour has $72\%$ of component $B$ ($y_B = 0.72$)! Why did this happen?

Because component $B$ has a higher pure vapour pressure ($12 	imes 10^3	ext{ Pa}$ vs $7 	imes 10^3	ext{ Pa}$), making it more volatile. The golden rule of solutions is that **the vapour phase is always richer in the more volatile component**. Keeping this conceptual check in mind can help you quickly verify your answers in exams!

Select Answer:

Visualized Solution

$x_{A} = 0.4, x_{B} = 0.6$

$p_{i} = x_{i} p_{i}^{\circ}$

$p_{A} = 0.4 \times 7 \times 1 0^{3}$

$p_{total} = p_{A} + p_{B}$

$y_{i} = \frac{p _{i}}{p _{total}}$

$y_{A} = 0.28, y_{B} = 0.72$

$y_{B} > x_{B}$

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

Analyzing the Setup

The Master Equation

Dalton's Law

The Volatility Rule

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