The vapour pressures of pure liquids A and B are 400 and 600 mmHg, respectively at 298 K. On mixing

Q: Why did we assume the solution is ideal?

The problem states that the sum of initial volumes equals the final volume ($\Delta V_{mix} = 0$), which is a primary characteristic of an ideal solution.

Q: What is the difference between $\chi_A$ and $Y_A$?

$\chi_A$ represents the mole fraction of component A in the liquid phase, while $Y_A$ represents its mole fraction in the vapour phase above the liquid.

Q: Why do we use Dalton's Law here?

Dalton's Law relates the partial pressure of a gas to its mole fraction in the gaseous mixture ($p_A = Y_A \cdot p_{total}$), allowing us to find the vapour phase composition.

Q: Can $Y_A$ and $\chi_A$ be equal?

Yes, but only if the pure vapour pressures of both components are exactly the same ($p_A^\circ = p_B^\circ$), which means they are equally volatile.

Q: What if the volumes didn't add up perfectly?

That would indicate a non-ideal solution with either positive or negative deviations from Raoult's Law, meaning the total pressure wouldn't simply be $p_A^\circ \chi_A + p_B^\circ \chi_B$.

Question

The vapour pressures of pure liquids $A$ and $B$ are $400$ and $600	ext{ mmHg}$, respectively at $298	ext{ K}$. On mixing the two liquids, the sum of their initial volumes is equal to the volume of the final mixture. The mole fraction of liquid $B$ is $0.5$ in the mixture. The vapour pressure of the final solution, the mole fractions of components $A$ and $B$ in vapour phase, respectively are

Options:

(A) $450	ext{ mmHg}, 0.4, 0.6$

(B) $500	ext{ mmHg}, 0.5, 0.5$

(C) $450	ext{ mmHg}, 0.5, 0.5$

(D) $500	ext{ mmHg}, 0.4, 0.6$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are looking at a beaker containing a mixture of two volatile liquids, A and B. The problem gives us a very crucial hint right at the beginning: it states that the sum of the initial volumes is exactly equal to the volume of the final mixture.

What does this tell us physically? It means that $\Delta V_{	ext{mix}} = 0$. There is no expansion or contraction upon mixing, which is the hallmark of an **ideal solution**. Because it is an ideal solution, we can confidently apply Raoult's Law without worrying about any positive or negative deviations.

### The Master Equation: Raoult's Law

According to Raoult's Law, the total vapour pressure of an ideal solution is simply the sum of the partial pressures of its individual components.

$$p_{	ext{total}} = p_A + p_B$$

We also know that the partial pressure of each component is the product of its pure vapour pressure and its mole fraction in the liquid phase.

$$p_{	ext{total}} = p_A^\circ \chi_A + p_B^\circ \chi_B$$

We are given the mole fraction of liquid B in the mixture as $\chi_B = 0.5$. Since the sum of all mole fractions in a mixture must always equal exactly 1, finding the mole fraction of liquid A is straightforward.

$$\chi_A = 1 - \chi_B = 1 - 0.5 = 0.5$$

### Calculating Total Vapour Pressure

Let's bring those numbers into our equation. We substitute the pure vapour pressures ($p_A^\circ = 400 	ext{ mmHg}$ and $p_B^\circ = 600 	ext{ mmHg}$) and the mole fractions we just found.

$$p_{	ext{total}} = 400(0.5) + 600(0.5)$$

Calculating this gives us the individual partial pressures: $p_A = 200 	ext{ mmHg}$ and $p_B = 300 	ext{ mmHg}$. Adding them together results in our total vapour pressure.

$$p_{	ext{total}} = 200 + 300 = \mathbf{500 	ext{ mmHg}}$$

### Vapour Phase Composition: Dalton's Law

We have the total pressure, but the question also asks for the composition in the vapour phase. For this, we shift our focus from Raoult's Law to Dalton's Law of partial pressures.

Dalton's Law tells us that the mole fraction of a component in the vapour phase (let's call it $Y_A$) is the ratio of its partial pressure to the total pressure.

$$Y_A = \frac{p_A}{p_{	ext{total}}}$$

Let's apply Dalton's Law for component A. The mole fraction of A in the vapour phase is its partial pressure ($200 	ext{ mmHg}$) divided by the total pressure ($500 	ext{ mmHg}$).

$$Y_A = \frac{200}{500} = \mathbf{0.4}$$

Finally, just like in the liquid phase, the mole fractions in the vapour phase must also add up to 1. So, the mole fraction of B in the vapour phase is simply 1 minus the mole fraction of A.

$$Y_B = 1 - Y_A = 1 - 0.4 = \mathbf{0.6}$$

Looking at our calculated values—a total pressure of $500 	ext{ mmHg}$, and vapour phase mole fractions of $0.4$ and $0.6$—we can see this perfectly matches option (d).

Select Answer:

Visualized Solution

Ideal Solution Setup

Raoult's Law

Mole Fractions in Liquid

Total Vapour Pressure

Dalton's Law

Mole Fraction of A in Vapour

Mole Fraction of B in Vapour

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

Analyzing the Setup

The Master Equation

Calculating Total Vapour Pressure

Vapour Phase Composition

Similar Questions

Comprehension Passage

At $300 K$ , the ratio of the molar volume of pure B in vapour phase to its molar volume in liquid phase is _____.

The mole fraction of B in vapour phase which is in equilibrium with this solution is ____.

The vapour pressures of $A$ and $B$ at $2 5^{\circ} C$ are $90 mm Hg$ and $15 mm Hg$ respectively. If $A$ and $B$ are mixed such that the mole-fraction of $A$ in the mixture is $0.6$ , then the mole fraction of $B$ in the vapour phase is $x \times 1 0^{- 1}$ . The value of $x$ is ......... .

Liquids A and B form ideal solution for all compositions of A and B at 25°C. Two such solutions with 0.25 and 0.50 mole fractions of A have the total vapor pressures of 0.3 and 0.4 bar, respectively. What is the vapor pressure of pure liquid B in bar?

A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of $290 mm$ at $300 K$ . The vapour pressure of propyl alcohol is $200 mm$ . If the mole fraction of ethyl alcohol is $0.6$ , its vapour pressure (in mm) at the same temperature will be

Benzene and toluene form nearly ideal solutions. At $2 0^{\circ} C$ , the vapour pressure of benzene is $75 torr$ and that of toluene is $22 torr$ . The partial vapour pressure of benzene at $2 0^{\circ} C$ for a solution containing $78 g$ of benzene and $46 g$ of toluene in torr is