Question

The vapour pressures of pure liquids A and B are 400 and 600 mm Hg, respectively at 298 K On mixing the two liquids, the sum of their initial volumes is equal to the volume of the final mixture. The mole fraction of liquids B is 0.5 in the mixture. The vapour pressure of the final solution, the mole fractions of components A and B in vapour phase, respectively are ______.

Options

• 450 mm Hg, 0.4, 0.6

• 500 mm Hg 0.5, 0.5

• 450 mm Hg, 0.5, 0.5

• 500 mm Hg, 0.4, 0.6

Answer 1

The vapour pressures of pure liquids A and B are 400 and 600 mm Hg, respectively at 298 K On mixing the two liquids, the sum of their initial volumes is equal to the volume of the final mixture. The mole fraction of liquids B is 0.5 in the mixture. The vapour pressure of the final solution, the mole fractions of components A and B in vapour phase, respectively are 500 mm Hg, 0.4, 0.6.

Explanation:

For ideal solution,

`"P"_"total" = "P"_"A"^circ xx "A" + "P"_"B"^circ xx "B"`

`"P"_"total" = {400 xx 0.5 + 600(1 - 0.5)}` mm Hg

= (200 + 300) mm Hg

= 500 mm Hg

Therefore mole fraction of A in vapour phase

`"Y"_"A" = "P"_"A"/"P"_"total" = ("P"_"A"^circ xx "A")/"P"_"total"`

`= (0.5 xx 400)/500`

= 0..4

Mole fraction of B in vapour phase,

Y_A + Y_B = 1

⇒ Y_B = 1 - Y_A

⇒ Y_B = 1 - 0.4

⇒ Y_B = 0.6