Two liquids A and B form an ideal binary solution. At , the total vapour pressure of a solution cont — Solutions and Colligative Properties Chemistry Question
Question
Two liquids A and B form an ideal binary solution. At $300\text{ K}$, the total vapour pressure of a solution containing $1\text{ mole}$ of A and $3\text{ moles}$ of B is $550\text{ mm of Hg}$. At the same temperature, if one more mole of B is added, the vapour pressure increases by $10\text{ mm of Hg}$. Determine the vapour pressure of pure A ($P_A^0$) in mm of Hg.
💡 Solution & Explanation
For Case I: $P_{total} = P_A^0 X_A + P_B^0 X_B \implies 550 = P_A^0(1/4) + P_B^0(3/4) \implies P_A^0 + 3P_B^0 = 2200$. For Case II, one more mole of B is added (total moles = 5), and $P_{total} = 560\text{ mm Hg}$. $560 = P_A^0(1/5) + P_B^0(4/5) \implies P_A^0 + 4P_B^0 = 2800$. Subtracting Case I from Case II yields $P_B^0 = 600\text{ mm Hg}$. Substituting back into Case I: $P_A^0 + 1800 = 2200 \implies P_A^0 = 400\text{ mm Hg}$.