Question

2 moles of liquid A and 3 moles of liquid B are mixed to from an ideal solution. The vapour pressure of ideal solution is 320 mm Hg. When 1 mole of A & 1 mole of B is further added then new vapour pressure of solution is 328.57 mm Hg. Find the vapour pressure of pure A ($P_A^\circ$) & pure B ($P_B^\circ$):

• A. $P_A^\circ = 200, P_B^\circ = 500$
• B. $P_A^\circ = 500, P_B^\circ = 200$
• C. $P_A^\circ = 300, P_B^\circ = 400$
• D. $P_A^\circ = 200, P_B^\circ = 300$

Hint:

Recognize decimal values like $.57$ as fractions ($4/7$) to facilitate integer arithmetic.

Answer 1

The Correct Option is B

To find the vapour pressures of pure components A and B (\(P_A^\circ\) and \(P_B^\circ\)), we will use Raoult's Law for ideal solutions. Raoult's Law states that:

\(P_{\text{solution}} = x_A \cdot P_A^\circ + x_B \cdot P_B^\circ\)

where \(x_A\) and \(x_B\) are the mole fractions of components A and B, respectively. Let's solve the problem step-by-step.

1. Initial solution: 2 moles of A and 3 moles of B.
   • Total moles = 2 + 3 = 5
   • Mole fraction of A, \(x_A = \frac{2}{5}\)
   • Mole fraction of B, \(x_B = \frac{3}{5}\)
   • Vapour pressure of this solution is given as 320 mm Hg.
2. Applying Raoult's Law to the initial solution:
   • \(320 = \left(\frac{2}{5}\right)P_A^\circ + \left(\frac{3}{5}\right)P_B^\circ\) (Equation 1)
3. New solution: 1 mole of A and 1 mole of B is added.
   • Total moles = 2 + 1 + 3 + 1 = 7
   • New mole fraction of A, \(x_A = \frac{3}{7}\)
   • New mole fraction of B, \(x_B = \frac{4}{7}\)
   • Vapour pressure of this solution is given as 328.57 mm Hg.
4. Applying Raoult's Law to the new solution:
   • \(328.57 = \left(\frac{3}{7}\right)P_A^\circ + \left(\frac{4}{7}\right)P_B^\circ\) (Equation 2)

We now have a system of two equations:

• \(320 = \frac{2}{5}P_A^\circ + \frac{3}{5}P_B^\circ\)
• \(328.57 = \frac{3}{7}P_A^\circ + \frac{4}{7}P_B^\circ\)

Solving these two equations simultaneously:

 

From Equation 1:

\(320 \times 5 = 2P_A^\circ + 3P_B^\circ\)
\(1600 = 2P_A^\circ + 3P_B^\circ\) (Multiply by 5 to eliminate the fraction)

From Equation 2:

\(328.57 \times 7 = 3P_A^\circ + 4P_B^\circ\)
\(2299.99 = 3P_A^\circ + 4P_B^\circ\) (Approximately, after multiplying by 7)

Using the elimination method, multiply Equation 1 by 3 and Equation 2 by 2 to eliminate \(P_A^\circ\):

\(4800 = 6P_A^\circ + 9P_B^\circ\) (Equation 3)
\(4599.98 = 6P_A^\circ + 8P_B^\circ\) (Equation 4)

Subtract Equation 4 from Equation 3:

\(4800 - 4599.98 = 1P_B^\circ\)
\(200.02 \approx 1P_B^\circ\)
\(P_B^\circ = 200\)

Substitute \(P_B^\circ = 200\) into Equation 1:

\(1600 = 2P_A^\circ + 3 \times 200\)
\(1600 = 2P_A^\circ + 600\)
\(1000 = 2P_A^\circ\)
\(P_A^\circ = 500\)

Thus, the vapour pressures of pure A and B are:

• \(P_A^\circ = 500 \, \text{mm Hg}\)
• \(P_B^\circ = 200 \, \text{mm Hg}\)

The correct answer is \(P_A^\circ = 500, P_B^\circ = 200\).