Question

100 g of liquid a (molar mass 140 g mor') was dissolved in 1000 g of liquid b (molar mass 180 g mol-'). 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. 63 se​

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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

Answer:

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

Now ptotal  = pA + pBOr ptotal = p°AXA + p°BxB

OR

475 = p°A X 0.114 + 500 X 0.886

475 = p°A X 0.114 + 500 X 0.886OR

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

NowpA = p°AXA

Or

p°A = pA / XA

⇒ 32 / 0.114

= 280.7 torr