Two liquids X and Y form an ideal solution at 300 K, vapour pressure of the solution containing 1 mo

Q: Why did the total moles change in the second case?

Because 1 mole of liquid Y was added to the solution, increasing the total number of moles from 4 to 5. This changes the mole fractions of both components.

Q: What does $P^\circ$ represent?

It represents the vapour pressure of the pure liquid at that specific temperature. It is a constant value for a given liquid at a constant temperature.

Q: Why do we multiply the equations by 4 and 5?

To clear the fractions! Multiplying by the total number of moles eliminates the denominators, giving us a much simpler system of linear equations to solve.

Q: Can we solve this using substitution instead of elimination?

Absolutely! You can isolate $P_X^\circ$ in the first equation ($P_X^\circ = 2200 - 3P_Y^\circ$) and substitute it into the second equation. You will get the exact same result.

Q: Does temperature affect these values?

Yes, vapour pressure is highly dependent on temperature. However, the problem explicitly states that the temperature remains constant at 300 K, so $P_X^\circ$ and $P_Y^\circ$ remain constant throughout the process.

Question

Two liquids $X$ and $Y$ form an ideal solution at $300	ext{ K}$, vapour pressure of the solution containing $1	ext{ mol}$ of $X$ and $3	ext{ mol}$ of $Y$ is $550	ext{ mmHg}$. At the same temperature, if $1	ext{ mol}$ of $Y$ is further added to this solution, vapour pressure of the solution increases by $10	ext{ mm Hg}$. Vapour pressure (in $	ext{mmHg}$) of $X$ and $Y$ in their pure states will be, respectively

Options:

(A) $200$ and $300$

(B) $300$ and $400$

(C) $400$ and $600$

(D) $500$ and $600$

Sigma Percentile · Accepted Answer

### The Magic of Raoult's Law

Imagine you are a detective, and you are given two mysterious beakers. In the first beaker, you have a mixture of $1	ext{ mol}$ of liquid $X$ and $3	ext{ mol}$ of liquid $Y$. The total vapour pressure hovering above this liquid is $550	ext{ mmHg}$.

In the second beaker, you decide to add just one more mole of liquid $Y$, bringing its total to $4	ext{ mol}$. Suddenly, the pressure gauge jumps to $560	ext{ mmHg}$. Your mission? Find the pure vapour pressures of $X$ and $Y$ ($P_X^\circ$ and $P_Y^\circ$).

### Setting Up the Equations

To solve this mystery, we need our trusty tool: **Raoult's Law**. It tells us that the total vapour pressure of an ideal solution is the sum of the partial pressures of its components. Mathematically, it looks like this:

$$P_{total} = P_X^\circ \chi_X + P_Y^\circ \chi_Y$$

Let's look at the first beaker. The total number of moles is $1 + 3 = 4$. So, the mole fraction of $X$ ($\chi_X$) is $\frac{1}{4}$, and the mole fraction of $Y$ ($\chi_Y$) is $\frac{3}{4}$. Plugging these into Raoult's Law gives us:

$$550 = P_X^\circ \left(\frac{1}{4}ight) + P_Y^\circ \left(\frac{3}{4}ight)$$

Nobody likes working with fractions, right? Let's multiply the entire equation by $4$ to clear the denominators. This gives us our first clean, linear equation:

$$P_X^\circ + 3P_Y^\circ = 2200 \quad \dots (i)$$

### The Second Beaker

Now, let's analyze the second beaker. We added $1	ext{ mol}$ of $Y$, so the total moles are now $1 + 4 = 5$. The new mole fractions are $\chi_X = \frac{1}{5}$ and $\chi_Y = \frac{4}{5}$. The new total pressure is $560	ext{ mmHg}$. Substituting these values yields:

$$560 = P_X^\circ \left(\frac{1}{5}ight) + P_Y^\circ \left(\frac{4}{5}ight)$$

Again, let's multiply the entire equation by $5$ to eliminate the fractions. This gives us our second linear equation:

$$P_X^\circ + 4P_Y^\circ = 2800 \quad \dots (ii)$$

### The Final Calculation

We now have a beautiful system of two linear equations.

$$P_X^\circ + 3P_Y^\circ = 2200$$
$$P_X^\circ + 4P_Y^\circ = 2800$$

If we subtract the first equation from the second, the $P_X^\circ$ terms cancel out perfectly!

$$(P_X^\circ + 4P_Y^\circ) - (P_X^\circ + 3P_Y^\circ) = 2800 - 2200$$

$$P_Y^\circ = 600	ext{ mmHg}$$

Now that we have the pure vapour pressure of $Y$, we can easily find $X$ by substituting $P_Y^\circ$ back into our first equation:

$$P_X^\circ + 3(600) = 2200$$

$$P_X^\circ + 1800 = 2200$$

$$P_X^\circ = 400	ext{ mmHg}$$

And there you have it! The pure vapour pressures of liquids $X$ and $Y$ are **$400	ext{ mmHg}$** and **$600	ext{ mmHg}$**, respectively. A perfect blend of chemistry concepts and basic algebra!

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Visualized Solution

$P_{t o t a l}$

$Raoult’s Law$

$State 1 Equation$

$Simplifying State 1$

$State 2 Equation$

$Simplifying State 2$

$Solving for P_{Y}^{\circ}$

$Solving for P_{X}^{\circ}$

$Final Answer$

$The Way Forward$

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

The Magic of Raoult's Law

Setting Up the Equations

The Second Beaker

The Final Calculation

Similar Questions

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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 ____.

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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 ......... .

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Visualized Solution

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

.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; } State 1 Equation

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } State 2 Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying State 2

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

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

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

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