Problem 165 - A Complete Resource Book in Chemistry for JEE Main

Question

Two liquids \(\mathrm{X}\) and \(\mathrm{Y}\) form an ideal solution. At \(300 \mathrm{~K}\), vapour pressure of the solutions containing 1 mol of \(X\) and 3 mol of \(Y\) is \(550 \mathrm{mmHg}\). At the same temperature, if 1 mol of \(Y\) is further added to this solu-tion, vapour pressure of the solution increases by 10 \(\mathrm{mmHg}\). Vapour pressure (in \(\mathrm{mmHg}\) ) of \(\mathrm{X}\) and \(\mathrm{Y}\) in their pure states will be, respectively: [2009] (a) 300 and 400 (b) 400 and 600 (c) 500 and 600 (d) 200 and 300

Step-by-Step Solution

Answer

The vapour pressures of \(X\) and \(Y\) are 400 mmHg and 600 mmHg respectively.

Step 1: Initial Mole Calculation

Initially, the solution contains 1 mole of liquid \(X\) and 3 moles of liquid \(Y\). Hence, the total number of moles is \(1 + 3 = 4\).

Step 2: Initial Mole Fractions

The mole fraction of \(X\), \(x_X\), is \(\frac{1}{4}\) and the mole fraction of \(Y\), \(x_Y\), is \(\frac{3}{4}\).

Step 3: Initial Vapour Pressure Equation

According to Raoult's Law, the vapour pressure of the solution is given by: \[ P_{total} = x_X P_X^0 + x_Y P_Y^0 = 550 \mathrm{mmHg} \]Where \(P_X^0\) and \(P_Y^0\) are the vapour pressures of \(X\) and \(Y\) in pure states.

Step 4: After Addition of 1 mol Y

When 1 mole of \(Y\) is added, the new moles are: 1 mole \(X\) and 4 moles \(Y\), making a total of 5 moles.

Step 5: New Mole Fractions

The new mole fraction of \(X\) is \(\frac{1}{5}\) and of \(Y\) is \(\frac{4}{5}\).

Step 6: New Vapour Pressure Equation

The new vapour pressure becomes:\[ P_{total}' = x_X' P_X^0 + x_Y' P_Y^0 = 560 \mathrm{mmHg} \]Since the pressure increased by 10 mmHg, we use this equation along with the initial vapour pressure equation.

Step 7: Solve the System of Equations

From the two equations, 1. \(\frac{1}{4} P_X^0 + \frac{3}{4} P_Y^0 = 550\)2. \(\frac{1}{5} P_X^0 + \frac{4}{5} P_Y^0 = 560\)Multiply the equations to eliminate fractions and solve simultaneously to find \(P_X^0\) and \(P_Y^0\).

Step 8: Calculate Vapor Pressures

Multiply equation (1) by 5 and equation (2) by 4 to obtain:\[\begin{align*}5 \left( \frac{1}{4} P_X^0 + \frac{3}{4} P_Y^0 \right) &= 2750 \4 \left( \frac{1}{5} P_X^0 + \frac{4}{5} P_Y^0 \right) &= 2240\end{align*}\]Simplify and subtract to find \(P_X^0 = 400\) mmHg and plug back to find \(P_Y^0 = 600\) mmHg.

Key Concepts

Ideal Solution

An ideal solution is a concept in chemistry that describes a perfect blending of two substances where their interactions are simply additive. This means that when two liquids, like our liquids X and Y, are mixed, they follow the properties of ideal solutions. Initially, they don't engage in any unexpected interactions. For example, one liquid does not disrupt the molecular structure of the other.
This behavior makes it easier to apply mathematical equations, like Raoult's Law, to predict various properties, such as vapor pressure. Ideal solutions are rare in the real world, as they assume perfect mixing with no heat absorption or release and no volume change upon mixing.

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid at a given temperature. It's an essential concept because it determines if a liquid will evaporate easily. The higher the vapor pressure, the faster the liquid turns into vapor.
In our problem, the solution containing liquids X and Y exhibits specific vapor pressures. The initial combined vapor pressure of 550 mmHg implies how readily the combined liquids were evaporating. After adding more Y, the pressure increases to 560 mmHg, showing that adding more of Y increases the rate at which molecules escape into the vapor phase. Using Raoult's Law, we can figure out how each component contributes to this overall vapor pressure.

Mole Fraction

The mole fraction is a way of expressing concentration in a solution. It is the ratio of moles of a component to the total number of moles in the solution. It's useful in scenarios like this because it tells us about the proportion of each liquid in the solution.
Initially, the mole fractions for X and Y were calculated as \(\frac{1}{4}\) and \(\frac{3}{4}\) respectively, reflecting one mole of X and three moles of Y in the mixture. When another mole of Y was added, these fractions changed to \(\frac{1}{5}\) for X and \(\frac{4}{5}\) for Y. These fractions are critical in applying Raoult’s Law to calculate the vapor pressures of X and Y in their pure states.

Pure State Pressure

Pure state pressure, often denoted by \(P^0\), is the vapor pressure of a liquid when it's not mixed with anything else. It essentially represents the maximum vapor pressure achievable by a substance in its pure state at a given temperature.
In this exercise, you are tasked with finding the pure state pressures of X and Y. Initially, these were unknown, but using the idea that our solution follows Raoult's Law, these pressures can be calculated. Based on the equations set up from our vapor pressure calculations, the pure state pressures for X and Y were determined to be 400 mmHg and 600 mmHg, respectively. This means that Y, having a higher pure state pressure, tends to evaporate more readily than X.