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Question
The vapour pressures of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if the total vapour pressure is 600 mm Hg. Also, find the composition of the vapour phase.
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Solution
Given: `P_A^0` = 450 mm Hg
`P_B^0` = 700 mm Hg
PTotal = 600 mm Hg
xA = ?
Applying Raoult’s law,
PA = `x_A xx P_A^0`
PB = `x_B xx P_B^0`
= `(1 - x_A)P_B^0`
PTotal = `P_A + P_B`
= `x_A xx P_A^0 + (1 - x_A)P_B^0`
= `P_B^0 + (P_A^0 - P_B^0)x_A`
Substituting the given values, we get
600 = 700 + (450 − 700)xA or 250xA = 100
or xA = `100/250` = 0.40
Thus, the composition of the liquid mixture will be
xA (mole fraction of A) = 0.40
xB (mole fraction of B) = 1 − 0.40 = 0.60
Calculation of composition in the vapour phase:
PA = `x_A xx P_A^0`
= 0.40 × 450 mm Hg
= 180 mm Hg
PB = `x_B xx P_B^0`
= 0.60 × 700 mm Hg
= 420 mm Hg
Mole fraction of A in the vapour phase = `(P_A)/(P_A + P_B)`
= ` 180/(180+420)`
= `180/600`
= 0.30
Mole fraction of B in the vapour phase = 1 − 0.30 = 0.70
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