Question

The total pressure observed by mixing two liquids A and B is 350 , mm Hg when their mole fractions are 0.7 and 0.3 respectively The total pressure becomes 410 ,mm ,Hg if the mole fractions are changed to 0.2 and 0.8 respectively for A and B The vapour pressure of pure A is _________mm ,Hg (Nearest integer) Consider the liquids and solutions behave ideally.

Hint:

Use Raoult's law: \(P_{\text{total}} = P_A^0 X_A + P_B^0 X_B\) for ideal solutions. Solve simultaneous equations for unknown vapour pressures.

Answer 1

Correct Answer: 314

Let the vapour pressures of pure A and B be \(P_A^0\) and \(P_B^0\), respectively.
For the first mixture:
\[P_{\text{total}} = P_A^0 X_A + P_B^0 X_B,\]
where \(P_{\text{total}} = 350 \, \text{mm Hg}, X_A = 0.7, X_B = 0.3\). Substituting:
\[350 = P_A^0 \cdot 0.7 + P_B^0 \cdot 0.3 \quad \text{(i)}.\]
For the second mixture:
\[P_{\text{total}} = P_A^0 X_A + P_B^0 X_B,\]
where \(P_{\text{total}} = 410 \, \text{mm Hg}, X_A = 0.2, X_B = 0.8\). Substituting:
\[410 = P_A^0 \cdot 0.2 + P_B^0 \cdot 0.8 \quad \text{(ii)}.\]
Solving equations (i) and (ii):
\[P_A^0 \cdot 0.7 + P_B^0 \cdot 0.3 = 350, \quad P_A^0 \cdot 0.2 + P_B^0 \cdot 0.8 = 410.\]
From (i):
\[P_B^0 = \frac{350 - P_A^0 \cdot 0.7}{0.3}.\]
Substitute into (ii):
\[410 = P_A^0 \cdot 0.2 + \left( \frac{350 - P_A^0 \cdot 0.7}{0.3} \right) \cdot 0.8.\]
Simplify:
\[410 = P_A^0 \cdot 0.2 + \frac{280 - P_A^0 \cdot 0.56}{0.3}.\]
\[410 = P_A^0 \cdot 0.2 + \frac{280}{0.3} - \frac{P_A^0 \cdot 0.56}{0.3}.\]
\[410 = P_A^0 \cdot 0.2 + 933.33 - 1.87 P_A^0.\]
\[410 = 933.33 - 1.67 P_A^0.\]
\[P_A^0 = \frac{933.33 - 410}{1.67}.\]
\[P_A^0 = 314 \, \text{mm Hg}.\]