Question

At 300 K the vapour pressure of an ideal solution containing 1.0 mole each of volatile liquids X and Y is 1000 mm. Keeping the temperature constant, when 2.0 moles of liquid X are added to the solution, its vapour pressure increases by 200 mm. Calculate the vapour pressure of X and Y in their pure state.

• A. \( P_X^0 = 1400,\; P_Y^0 = 600 \)
• B. \( P_X^0 = 1700,\; P_Y^0 = 400 \)
• C. \( P_X^0 = 1200,\; P_Y^0 = 480 \)
• D. \( P_X^0 = 1000,\; P_Y^0 = 500 \)

Hint:

For ideal solutions, always use Raoult's law and form simultaneous equations based on mole fractions before and after change.

Answer 1

The Correct Option is A

Step 1: Apply Raoult's Law.
For an ideal solution:
\[ P = x_X P_X^0 + x_Y P_Y^0 \]

Step 2: Initial condition.
Moles of X = 1, moles of Y = 1
Total moles = 2
\[ x_X = x_Y = \frac{1}{2} \]
Given total pressure = 1000 mm
\[ \frac{1}{2}P_X^0 + \frac{1}{2}P_Y^0 = 1000 \]
\[ P_X^0 + P_Y^0 = 2000 \quad \text{(Equation 1)} \]

Step 3: After adding 2 moles of X.
New moles of X = 3, moles of Y = 1
Total moles = 4
\[ x_X = \frac{3}{4}, \quad x_Y = \frac{1}{4} \]
New vapour pressure = \( 1000 + 200 = 1200 \) mm
\[ \frac{3}{4}P_X^0 + \frac{1}{4}P_Y^0 = 1200 \]
Multiply by 4:
\[ 3P_X^0 + P_Y^0 = 4800 \quad \text{(Equation 2)} \]

Step 4: Solve equations.
From Equation 1:
\[ P_Y^0 = 2000 - P_X^0 \]
Substitute in Equation 2:
\[ 3P_X^0 + (2000 - P_X^0) = 4800 \]
\[ 2P_X^0 + 2000 = 4800 \]
\[ 2P_X^0 = 2800 \]
\[ P_X^0 = 1400 \]
\[ P_Y^0 = 2000 - 1400 = 600 \]

Step 5: Conclusion.
Thus, vapour pressures of pure liquids are:
\[ \boxed{P_X^0 = 1400,\; P_Y^0 = 600} \]