Liquids A and B form ideal solution over the entire range of composition. At temperature T, equimola

Q: Why did we assume $x_A = 0.5$ and $x_B = 0.5$ in the first case?

The problem explicitly states it is an "equimolar" binary solution. "Equi" means equal, and "molar" refers to moles. Therefore, the number of moles of A equals the number of moles of B, making their mole fractions exactly 0.5 each.

Q: Can we use Raoult's Law for any liquid mixture?

No, Raoult's Law in its simple form $P_T = P_A^\circ x_A + P_B^\circ x_B$ strictly applies only to *ideal* solutions. The problem specifically mentioned that liquids A and B form an ideal solution, which gave us the green light to use this equation.

Q: What if the vapor pressure of pure A wasn't given? Could we still solve it?

Without $P_A^\circ$, we would only know that $P_A^\circ + P_B^\circ = 90$. We would have two variables and only one equation, making it impossible to find the exact individual pure vapor pressures needed for the second part of the problem.

Q: Why did the total vapor pressure drop from 45 Torr to 22.5 Torr in the second case?

The total vapor pressure depends on the composition. We found that $P_A^\circ = 20$ Torr and $P_B^\circ = 70$ Torr. Liquid A is much less volatile than liquid B. The pressure dropped significantly because the new mixture is heavily dominated by the less volatile liquid A ($x_A = 0.95$).

Q: Is it possible for a mole fraction to be greater than 1 or negative?

Absolutely not. Mole fractions represent a part of a whole. By definition, a mole fraction must always be between 0 and 1, and the sum of all mole fractions in a mixture must exactly equal 1. If you ever calculate a negative mole fraction, check your algebra!

Question

Liquids A and B form ideal solution over the entire range of composition. At temperature T, equimolar binary solution of liquids A and B has vapour pressure 45 Torr. At the same temperature, a new solution of A and B having mole fractions $x_A$ and $x_B$, respectively, has vapour pressure of 22.5 Torr. The value of $x_A/x_B$ in the new solution is_______.
(Given that the vapour pressure of pure liquid A is 20 Torr at temperature T)

Sigma Percentile · Accepted Answer

### Setting the Stage: The Tale of Two Mixtures

Imagine stepping into a physical chemistry laboratory. On the bench in front of you sit two identical glass beakers, both maintained at a constant temperature, $T$.

These aren't just any liquids; they are liquids A and B, and they form a perfectly **ideal solution**. This means the molecules of A and B interact with each other exactly as they would with their own kind. There are no unexpected attractions or repulsions, making our mathematical journey beautifully predictable.

In the first beaker, we have an **equimolar mixture**. This is our first major clue. "Equimolar" means the number of moles of liquid A is exactly equal to the number of moles of liquid B.

Consequently, their mole fractions must be equal. Since the sum of mole fractions in a binary mixture is always $1$, we can confidently state that $x_A = 0.5$ and $x_B = 0.5$. The total vapor pressure hovering above this liquid mixture is measured to be $45 	ext{ Torr}$.

### The Master Equation: Raoult's Law in Action

To connect the macroscopic world of pressure to the microscopic world of mole fractions, we need a theoretical bridge. That bridge is **Raoult's Law**.

For an ideal binary solution, Raoult's Law states that the total vapor pressure is the sum of the partial vapor pressures of each component. Mathematically, this is expressed as:

$$P_{	ext{total}} = P_A^\circ x_A + P_B^\circ x_B$$

Here, $P_A^\circ$ and $P_B^\circ$ represent the vapor pressures of pure liquids A and B, respectively. Let's apply this master equation to our first beaker.

Substituting our known values, we get:

$$45 = P_A^\circ \left(\frac{1}{2}ight) + P_B^\circ \left(\frac{1}{2}ight)$$

This equation looks a bit cumbersome with the fractions. Let's multiply the entire equation by $2$ to clear the denominators. This yields a remarkably clean and simple relationship:

$$P_A^\circ + P_B^\circ = 90$$

### Unveiling the Pure Vapor Pressures

We have one equation but two unknown pure vapor pressures. However, the problem statement hands us a crucial piece of information on a silver platter.

We are explicitly given that the vapor pressure of pure liquid A at temperature $T$ is $20 	ext{ Torr}$. This means $P_A^\circ = 20 	ext{ Torr}$.

Let's substitute this gift into our simplified relationship:

$$20 + P_B^\circ = 90$$

Solving for $P_B^\circ$ is now a matter of simple arithmetic. Subtracting $20$ from $90$, we find:

$$P_B^\circ = 70 	ext{ Torr}$$

Take a moment to appreciate what this means physically. Liquid B has a pure vapor pressure of $70 	ext{ Torr}$, while liquid A is only $20 	ext{ Torr}$. Liquid B is significantly more volatile; it wants to escape into the vapor phase much more readily than liquid A.

### The Second Mixture: A Shift in Equilibrium

Now, let's shift our attention to the second beaker. The temperature remains the same, so our pure vapor pressures, $P_A^\circ$ and $P_B^\circ$, remain locked at $20 	ext{ Torr}$ and $70 	ext{ Torr}$.

However, the recipe has changed. We have a new solution with unknown mole fractions, $x_A$ and $x_B$. The total vapor pressure has plummeted to $22.5 	ext{ Torr}$.

Why did the pressure drop so drastically? Because the new mixture must be heavily dominated by the less volatile component, liquid A. Let's prove this mathematically by applying Raoult's Law once more:

$$22.5 = 20 x_A + 70 x_B$$

We are faced with one equation and two unknowns. But remember the golden rule of mole fractions: in a binary mixture, they must sum to exactly $1$.

Therefore, we can express $x_B$ in terms of $x_A$:

$$x_B = 1 - x_A$$

### The Final Calculation: Isolating the Mole Fractions

Let's substitute this expression for $x_B$ back into our Raoult's Law equation. This will give us an equation with only one variable, $x_A$:

$$22.5 = 20 x_A + 70 (1 - x_A)$$

Now, we carefully expand the brackets. Distributing the $70$ gives us:

$$22.5 = 20 x_A + 70 - 70 x_A$$

Combining the $x_A$ terms on the right side, we get:

$$22.5 = 70 - 50 x_A$$

Let's rearrange the terms to isolate $x_A$. Moving $50 x_A$ to the left and $22.5$ to the right yields:

$$50 x_A = 70 - 22.5$$

$$50 x_A = 47.5$$

Dividing both sides by $50$, we find the mole fraction of liquid A:

$$x_A = \frac{47.5}{50} = 0.95$$

As we suspected, the solution is $95\%$ liquid A! Now, finding $x_B$ is trivial:

$$x_B = 1 - 0.95 = 0.05$$

### The Grand Finale: The Ratio

We have successfully decoded the composition of the new solution. The final step is to answer the specific question asked: what is the value of the ratio $\frac{x_A}{x_B}$?

Let's plug in our calculated mole fractions:

$$\frac{x_A}{x_B} = \frac{0.95}{0.05}$$

Since both numbers have two decimal places, we can multiply the numerator and denominator by $100$ to simplify the fraction:

$$\frac{x_A}{x_B} = \frac{95}{5}$$

Performing the final division, we arrive at our destination:

$$\frac{x_A}{x_B} = 19$$

**The final answer is 19.** This problem is a beautiful demonstration of how macroscopic measurements, like total vapor pressure, can be used to precisely determine the microscopic composition of a mixture using the elegant framework of Raoult's Law.

Enter Numerical Value:

Visualized Solution

$V i s u a l i z in g t h e T w o S t a t es$

$R a o u l t^{'} s L a w$

$C a se 1 : E q u im o l a r S e t u p$

$S im pl i f y in g C a se 1$

$F in d in g P_{B}^{\circ}$

$C a l c u l a t in g P_{B}^{\circ}$

$C a se 2 : N e w S o l u t i o n S e t u p$

$E l imina t in g x_{B}$

$S o l v in g f or x_{A}$

$C a l c u l a t in g M o l e F r a c t i o n s$

$F ina l R a t i o$

$T h e W a y F or w a r d$

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

Setting the Stage

The Master Equation

Unveiling the Pure Vapor Pressures

The Second Mixture

The Final Calculation

The Grand Finale

Similar Questions

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?

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 ......... .

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

At $2 0^{\circ} C$ , the vapour pressure of benzene is 70 torr and that of methyl benzene is 20 torr. The mole fraction of benzene in the vapour phase at $2 0^{\circ} C$ above an equimolar mixture of benzene and methyl benzene is...... $\times 1 0^{- 2}$ . (Nearest integer)

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

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Two States

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Raoult′s Law

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case 1: Equimolar Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying Case 1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding PB∘​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating PB∘​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case 2: New Solution Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Eliminating xB​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for xA​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Mole Fractions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Ratio

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward