Question
At $T(K),2$ moles of liquid A and 3 moles of liquid B are mixed. The vapour pressure of ideal solution formed is 320 mm Hg . At this stage, one mole of A and one mole of B are added to the solution. The vapour pressure is now measured as 328.6 mm Hg . The vapour pressure (in mm Hg ) of A and B are respectively:
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Solution
2 moles of $A + 3$ moles of B
$$X_{A} = 2/5,X_{B} = 3/5$$
$${P_{S} = X_{A}P_{A}^{\circ} + X_{B}P_{B}^{\circ} }{320 = P_{A}^{\circ}\left( \frac{2}{5} \right) + P_{B}^{\circ}\left( \frac{3}{5} \right)}$$
$$\begin{array}{r} 2P_{A}^{\circ} + 3P_{B}^{\circ} = 1600\#(I) \end{array}$$
Now 1 mole of A & 1 mole of B is added
$${X_{A}' = \frac{3}{7},X_{B}' = \frac{4}{7} }{P_{S}' = 328.6 = P_{A}^{\circ}\left( \frac{3}{7} \right) + P_{B}^{\circ}\left( \frac{4}{7} \right)}$$
$$\begin{array}{r} 3P_{A}^{\circ} + 4P_{B}^{\circ} = 2300.2\#(II) \end{array}$$
Now eq (I) $\times 3 -$ eq (II) $\times 2$
$${6P_{A}^{\circ} + 9P_{B}^{\circ} = 4800 }{6P_{A}^{\circ} + 8P_{B}^{\circ} = 4600.4 }$$$P_{B}^{\circ} \simeq 200\text{ }mm$ of Hg
$P_{A}^{\circ} \simeq 500\text{ }mm$ of Hg
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