Two liquids A and B form an ideal solution. At 320 K, the vapour pressure of the solution, containing 3 mol of A and 1 mol of B is 500 mm Hg. At the same temperature, if 1 mol of A is further added to this solution, vapour pressure of the solution increas

Question

Two liquids A and B form an ideal solution. At 320 K, the vapour pressure of the solution, containing 3 mol of A and 1 mol of B is 500 mm Hg. At the same temperature, if 1 mol of A is further added to this solution, vapour pressure of the solution increases by 20 mm Hg. Vapour pressure (in mm Hg) of B in pure state is ______. (Nearest integer) [2026]

Smart Achievers · Accepted Answer

(200)

$X_{A} = \frac{3}{4}, X_{B} = \frac{1}{4}$

$P_{s} = P_{A} ° X_{A} + P_{B} ° X_{B}$

$500 = P_{A} ° \times \frac{3}{4} + P_{B} ° \times \frac{1}{4}$

$3 P_{A} ° + P_{B} ° = 2000 . . . . (1)$

$Now 1 mole of A is further added so n_{A} = 4 mole, n_{B} = 1 mole$

$X'_{A} = \frac{4}{5}, X'_{B} = \frac{1}{5}$

$P_{s} = 520 = P_{A} ° \times \frac{4}{5} + P_{B} ° \times \frac{1}{5}$

$4 P_{A} ° + P_{B} ° = 2600 . . . (2)$

$By equation (2) - equation (1)$

$P_{A} ° = 600 mm Hg$

$P_{B} ° = 200 mm Hg$

Solution

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Solution

Ans. (200) XA=34, XB=14 Ps=PA°XA+PB°XB 500=PA°×34+PB°×14 3PA°+PB°=2000 ....(1) Now 1 mole of A is further added so nA=4 mole, nB=1 mole X'A=45, X'B=15 Ps=520=PA°×45+PB°×15 4PA°+PB°=2600 ...(2) By equation (2) - equation (1) PA°=600 mm Hg PB°=200 mm Hg

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