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

To find the vapor pressures of the pure components X and Y, we can utilize Raoult's Law, which states that the vapor pressure of a solution is dependent on the vapor pressures of the pure components and their mole fractions in the solution. Let's break down the problem step by step.

Understanding the Initial Setup

We know the following:

• The solution at 300 K has 1 mol of X and 3 mol of Y, giving a total of 4 moles.
• The vapor pressure of this solution is 550 mmHg.

Calculating Mole Fractions

The mole fraction of each component can be calculated as follows:

• Mole fraction of X, \( X_X = \frac{1}{1 + 3} = \frac{1}{4} = 0.25 \)
• Mole fraction of Y, \( X_Y = \frac{3}{1 + 3} = \frac{3}{4} = 0.75 \)

Applying Raoult's Law

According to Raoult's Law, the vapor pressure of the solution (\( P_{solution} \)) can be expressed as:

P_solution = X_X * P^0_X + X_Y * P^0_Y

Where \( P^0_X \) and \( P^0_Y \) are the vapor pressures of pure X and Y, respectively. Plugging in the known values, we have:

550 mmHg = (0.25 * P^0_X) + (0.75 * P^0_Y)

After Adding More Y

Now, we add 1 mol of Y, which makes the new composition:

• 1 mol of X
• 4 mol of Y

This gives a total of 5 moles. The new mole fractions are:

• Mole fraction of X, \( X_X = \frac{1}{1 + 4} = \frac{1}{5} = 0.20 \)
• Mole fraction of Y, \( X_Y = \frac{4}{1 + 4} = \frac{4}{5} = 0.80 \)

Using Raoult's Law again for the new solution, we have:

P_new = X_X * P^0_X + X_Y * P^0_Y

Since the vapor pressure increases by 10 mmHg, the new pressure is:

P_new = 550 mmHg + 10 mmHg = 560 mmHg

Substituting the mole fractions into the equation gives:

560 mmHg = (0.20 * P^0_X) + (0.80 * P^0_Y)

Setting Up Equations

Now we have a system of two equations:

• Equation 1: \( 550 = 0.25 * P^0_X + 0.75 * P^0_Y \)
• Equation 2: \( 560 = 0.20 * P^0_X + 0.80 * P^0_Y \)

Solving the Equations

We can solve these equations simultaneously. Start with Equation 1:

0.25 * P^0_X + 0.75 * P^0_Y = 550

From this, we can express \( P^0_Y \) in terms of \( P^0_X \):

P^0_Y = \frac{550 - 0.25 * P^0_X}{0.75}

Now substituting this expression into Equation 2:

560 = 0.20 * P^0_X + 0.80 * \left(\frac{550 - 0.25 * P^0_X}{0.75}\right)

Multiplying through by 0.75 to eliminate the fraction:

420 = 0.15 * P^0_X + 0.80 * (550 - 0.25 * P^0_X)

Expanding and simplifying gives:

420 = 0.15 * P^0_X + 440 - 0.20 * P^0_X

Combining like terms results in:

420 - 440 = -0.05 * P^0_X

-20 = -0.05 * P^0_X

Thus, we find:

P^0_X = 400 mmHg

Finding the Vapor Pressure of Y

Now substituting \( P^0_X \) back into the expression for \( P^0_Y \):

P^0_Y = \frac{550 - 0.25 * 400}{0.75} = \frac{550 - 100}{0.75} = \frac{450}{0.75} = 600 mmHg

Final Result

In conclusion, the vapor pressures of the pure components are:

• P^0_X = 400 mmHg
• P^0_Y = 600 mmHg