Raoult's law - Vapor pressure

Wek

The density of toluene is 0.862g/ml at 25C and it's vapor pressure s 28.44 mmHg. At 25C, the vapor pressure of toluene above a solution of C10H8 in 500ml toluene is 27.92 mmHg. How many grams of C10H8 are present in the solution?

I first used Raoult's equation to find X. Then I used the density and given volume of toluene to find the grams and then the moles of it. Then using the moles fraction equation I found the moles of C10H8 and using the FW of it I found the grams of it.

I did the work and come up with 12.8g, which is wrong. The book says it's 8.03g.

Can anyone tell me what I might have done wrong? Thanks

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Borek

Mentor

I didn't get neither 12.8g nor 8.03g Perhaps it is too early.

Wek

This is what I did:

-To find mole fraction of toluene I divided the vapor pressure of toluene solution by toluene's vapor pressure: 27.92/28.44 = 0.9817
-To get the moles of toluene: density x toluene volume then divided by toluene FW: 431g/92.0g = 4.68mol
-Using the moles fraction equation I found the moles of C10H18: C10H18 mol = (4.68/0.9817) - 4.68 = 0.10mol
-To get the mass of C10H18 I multiplied the moles of C10H18 by its FW: 0.10mol x 128g = 12.8g

Borek

Mentor
1. Are you rounding down intermediate results? If so - don't.

2. C10H8 or C10H18?

C10H18: C10H18 mol = (4.68/0.9817) - 4.68 = 0.10mol

You are not getting 8.03g, but (after taking into account math errors) you are very close to my result. Either we are both wrong, or the book is wrong. I tend to think we are right.

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