Question

100 g of liquid A (molar mass 140 g mol-1) was dissolved in 1000 g of liquid B (molar mass 180 g mol-1). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.

Answer 1

Answer:

Explanation:

Number of Moles of Liquid A, nA = 100 / 140 = 0.714

Number of Moles of Liquid B, nB = 1000 / 180 = 5.556

Then Mole fraction of A = nA / nA + nB = 0.714 / 0.714 + 5.556

= 0.114

Now Mole of fraction of B = 1 - 0.114 = 0.886

Now ptotal  = pA + pB

Or ptotal = p°AXA + p°BxB

OR

475 = p°A X 0.114 + 500 X 0.886

OR

p°A = 280.7 torr

Therefore vapour pressure of pure A = 280.7 torr

Vapour pressure of A in solution = 280.7 x 0.114

= 32 torr

Now

pA  = p°AXA

Or

p°A  = pA   /  XA

⇒ 32 / 0.114

= 280.7 torr

Hence, the vapour pressure of pure liquid A is 280.7 torr.

Answer 2

Explanation:

Number of Moles of Liquid A, nA = 100 / 140 = 0.714

Number of Moles of Liquid B, nB = 1000 / 180 = 5.556

Then Mole fraction of A = nA / nA + nB = 0.714 / 0.714 + 5.556

= 0.114

Now Mole of fraction of B = 1 - 0.114 = 0.886

Now ptotal = pA + pB

Or ptotal = p°AXA + p°BxB

OR

475 = p°A X 0.114 + 500 X 0.886

OR

p°A = 280.7 torr

Therefore vapour pressure of pure A = 280.7 torr

Vapour pressure of A in solution = 280.7 x 0.114

= 32 torr

Now

pA = p°AXA

Or

p°A = pA / XA

⇒ 32 / 0.114

= 280.7 torr

Hence, the vapour pressure of pure liquid A is 280.7 torr.

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