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??????

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# Phase equilibrium

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