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?(adsbygoogle = window.adsbygoogle || []).push({});

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??????

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Phase equilibrium

**Physics Forums | Science Articles, Homework Help, Discussion**