Phase equilibrium

  • Thread starter siewwen168
  • Start date
15
0
The vapour pressure of a pure liquid solvent B is 75 kPa. When an unknown mass of non volatile solute W was added to 0.2 moles of solvent B,its vapour pressure drops by 20 kPa. If the relative molecular mass of W is 118,how much of W was added to B?

A 5.07g
B 8.58g
C 23.44g
D 64.9g

i will show my calculation here,but i don't think its correct because my answer wrong .please state out my mistake.

P(B) = X(B) x P(pure B)
20 = (0.2/W+0.2) x 75
W =0.55

mass of W = 0.55x118
= 64.90g
the actual answer is B,i think my answer is unreasonable too,cause the mass is too big,but from my calculation,i get 64.9g.why??????
 

GCT

Science Advisor
Homework Helper
1,727
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20 = (0.2/W+0.2) x 75
wrong value for the vapor pressure, it has dropped by 20kPa
 

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