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Question

100 g of liquid A (molar mass 140 g mol1) was dissolved in 1000 g of liquid B (molar mass 180 g mol1) 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:

  1. 500,300
  2. 190,31.965
  3. 3.19,500
  4. 283.18,31.999

A
283.18,31.999
B
3.19,500
C
500,300
D
190,31.965
Solution
Verified by Toppr

Given: PT=475,WA=100g,MA=140,PoB=500
WB=1000g,MB=180,PoA?,PA=?
PT=PoAXA+PoBXB(1) & XA+XB=1(2)
XBXA=WBMAWAMB=1000×140100×180=7.78
XB=7.78XA
Substituting XB in (2)
XA+7.78XA=1
XA=0.113 & hence XB=10.113=0.886
Substituting XA & XB in equation (1)
475=PoA(0.113)+500(0.886)
0.113PoA=32PoA=320.113=283.18torr
Now, PA=PoA(1XB)=PoAXA
=283.18×0.113
=31.999torr

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