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??????