The vapour pressures of pure liquids A B are 450 mm and 700 mm of Hg respectively at 350 K. Calculate the compositon of the liquid mixture if total vapour pressure is 600 mm of Hg. Also find the composition in the Vapour phase.

Answer

0.4,0.6,0.3,0.7

Explanation

Let mole fraction of liquid A in solution = `x_(A)`
Mole fraction of liquid B in solution , `x_(B) = 1 - x_(A)`
` p = p_(A)^(@)x_(A) + p_(B)^(@)x_(B) or = p_(A)^(@) x_(A) + p_(B)^(@) (1-x_(A))`
` p = 600mm Hg`
` 600 = 450 x_(A) = 700(1-x_(A))`
solving `" " x_(A) = (100)/(250) = 0.4`
Mole fraction of liquid A = 0.4
Mole fraction of liquid B = 1 -0.4 = 0.6
Calculation of composition of vapour phase .
`p_(A) = p_(A)^(@)x_(A) = 450 mm xx 0.4 = 180`
`p_(B) = p_(B)^(@) x_(B) = 700 xx 0.6 = 420 mm`
`p_("total") = p_(A) + p_(B) = 180+420 = 600 mm `
`y_(A) = (p_(A))/(p_("total")) = (180)/(600) = 0.3`
`y_(B) = (p_(B))/(p_("total")) = (420)/(600) = 0.7`