The vapour pressures of A and B at 25°C are 90 mm Hg and 15 mm Hg respectively. If A and B are mixed

Q: Why is the mole fraction of B in the vapour phase different from its mole fraction in the liquid phase?

The vapour phase is always richer in the more volatile component. Since pure A has a much higher vapour pressure ($90 \text{ mm Hg}$) than pure B ($15 \text{ mm Hg}$), A evaporates more readily. Even though A is $60\%$ of the liquid, it makes up $90\%$ of the vapour ($54/60$). Consequently, B's mole fraction drops from $0.4$ in the liquid to $0.1$ in the vapour.

Q: What is the difference between $\chi_B$ and $Y_B$?

$\chi_B$ represents the mole fraction of component B in the **liquid solution**. $Y_B$ represents the mole fraction of component B in the **gaseous vapour mixture** above the liquid. They are calculated using different laws (Raoult's vs. Dalton's).

Q: Can I just use the ratio of their pure vapour pressures to find the vapour phase mole fraction?

No, that would only work if the liquid mixture was exactly equimolar ($\chi_A = \chi_B = 0.5$). The actual partial pressure depends on *both* the pure vapour pressure and how much of that liquid is present (its mole fraction).

Q: What if the solution was non-ideal?

If the solution showed positive or negative deviations from Raoult's Law, we would need to know the activity coefficients to calculate the true partial pressures. Since no such data is given, we assume ideal behavior.

Q: Why do we add the partial pressures to get the total pressure?

This is based on Dalton's Law of Partial Pressures, which states that in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of the individual gases.

Question

The vapour pressures of $A$ and $B$ at $25^{\circ}	ext{C}$ are $90	ext{ mm Hg}$ and $15	ext{ mm Hg}$ 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 $x 	imes 10^{-1}$. The value of $x$ is ......... .

Sigma Percentile · Accepted Answer

Imagine you are looking at a closed container holding a mixture of two volatile liquids, $A$ and $B$. The space above the liquid isn't empty; it's filled with the vapours of both $A$ and $B$ constantly evaporating and condensing. This dynamic equilibrium is governed by two beautiful laws of physical chemistry: **Raoult's Law** and **Dalton's Law of Partial Pressures**.

### Analyzing the Liquid Phase

We are given the pure vapour pressures of the two liquids at $25^{\circ}	ext{C}$: $p_A^\circ = 90 	ext{ mm Hg}$ and $p_B^\circ = 15 	ext{ mm Hg}$. We also know the composition of the liquid mixture. The mole fraction of $A$ in the liquid phase is $\chi_A = 0.6$.

Since a mixture is made entirely of its components, the sum of all mole fractions must equal $1$. Therefore, the mole fraction of $B$ in the liquid phase is simply:

$$\chi_B = 1 - \chi_A = 1 - 0.6 = 0.4$$

### Applying Raoult's Law

Raoult's Law tells us that the partial vapour pressure of a component in an ideal solution is directly proportional to its mole fraction in the liquid phase. Let's calculate the partial pressures $p_A$ and $p_B$ exerted by the vapours of $A$ and $B$ respectively.

$$p_A = \chi_A p_A^\circ = 0.6 	imes 90 = 54 	ext{ mm Hg}$$

$$p_B = \chi_B p_B^\circ = 0.4 	imes 15 = 6 	ext{ mm Hg}$$

### The Total Vapour Pressure

Now, we look at the vapour phase as a whole. According to Dalton's Law of Partial Pressures, the total pressure $p_T$ exerted by a mixture of non-reacting gases is the sum of their individual partial pressures.

$$p_T = p_A + p_B = 54 + 6 = 60 	ext{ mm Hg}$$

### Composition of the Vapour Phase

The question asks for the mole fraction of $B$ in the vapour phase, which we will denote as $Y_B$. Dalton's Law also provides a direct relationship between the partial pressure of a gas, the total pressure, and its mole fraction in the gaseous mixture: $p_B = Y_B p_T$.

Rearranging this, we can find $Y_B$:

$$Y_B = \frac{p_B}{p_T} = \frac{6}{60} = 0.1$$

### Final Calculation

The problem states that the mole fraction of $B$ in the vapour phase is given in the format $x 	imes 10^{-1}$. Let's convert our result into scientific notation to match this format.

$$Y_B = 0.1 = 1 	imes 10^{-1}$$

By comparing $1 	imes 10^{-1}$ with $x 	imes 10^{-1}$, it is crystal clear that the value of $x$ is exactly **$1$**.

Enter Numerical Value:

Visualized Solution

\text{Liquid Mixture Setup}

\text{Raoult's Law}

\text{Partial Pressures}

\text{Total Vapour Pressure}

\text{Mole Fraction in Vapour Phase}

\text{Calculating } Y_B

\text{Comparing with Given Format}

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

Analyzing the Liquid Phase

Applying Raoult's Law

The Total Vapour Pressure

Composition of the Vapour Phase

Final Calculation

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