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

[siewwen168]

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]

20 = (0.2/W+0.2) x 75

wrong value for the vapor pressure, it has dropped by 20kPa

 

[ravenprp]

After reviewing your calculation, it appears that your mistake is in the first equation. The correct equation should be:

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

Solving for W, we get:
W = 0.44 moles

To convert moles to grams, we use the molar mass of W (118 g/mol):
Mass of W = 0.44 mol x 118 g/mol = 52.05 g

Therefore, the correct answer is not any of the given options. It should be 52.05g. Please double check your calculations to ensure accuracy.