Calculating Mass of Non-Volatile Solute W in Solvent B for Phase Equilibrium

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The discussion centers on calculating the mass of a non-volatile solute W added to a pure liquid solvent B, which has a vapour pressure of 75 kPa. When 0.2 moles of solvent B is mixed with solute W, the vapour pressure decreases by 20 kPa. The calculation attempts to determine the mass of W, given its relative molecular mass of 118. The initial calculation incorrectly uses the formula for vapour pressure, leading to an erroneous result of 64.9g for the mass of W. The mistake lies in the application of the vapour pressure change; the correct approach should account for the new vapour pressure after the decrease, which is 55 kPa. The correct calculation should yield a mass of 8.58g, aligning with the expected answer. The discussion emphasizes the importance of accurately applying the vapour pressure formula to avoid significant discrepancies in results.
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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?
 
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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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