# Raoult's law - Vapor pressure

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

## Answers and Replies

Borek
Mentor
Sounds like your approach is a correct one, show your numbers.

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

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

Check your math.

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.