Vapour pressure problem (simple)

1. Jun 9, 2009

geffman1

The vapour pressure of pure chloroform at 70.0oC is 135.7 kPa. What mass of iodine (I2) should be dissolved in 1 L of chloroform, CHCl3 (density = 1.49 g/cm3) to lower the vapour pressure by 13.3 kPa?

hey guys i really cant do this question, i think in involves raults law, P=XPA
anyhelp would be good. thanks

p.s. answer is 1.32 x 101 g

2. Jun 10, 2009

Staff: Mentor

No idea if you are to take into account iodine pressure or not.

Assuming it has to be ignored - at what x the pressure of chloroform vapour will be lowered by 13.3 kPa?

3. Jun 10, 2009

geffman1

thnks for the reply, umm 122.4kpa

4. Jun 17, 2009

bluesman4509

Hey,

Here is the solution I got, its different to your answer though...

Psolution=xsolvent.Psolvent

so Psolution=135.7-13.3 kpa (the new pressure of the solution)

now in 1L =1000cm3 there is 1490 grams = 12.482 moles

so re-arrange the equation and you get=> x = (12.482 - 11.26)/(3.55x10-3)

=344.23 grams

if 133.32 is the correct answer as you suggest I dont know where went wrong.

5. Jun 17, 2009

geffman1

hey mate i checked with the teacher and ur right(stupid internet tests). just wondering if you could show me where you got a few of the those numbers from, ive tried but get lost. thanks

6. Jun 20, 2009

bluesman4509

Ok,

so Xsolvent=molsolvent/(molsolute+molsolvent), we are looking to find molsolute (I2) so we can multiply by the RFM (253.8 g.mol-1) to get the mass needed.

1L of chloroform has mass 1490g = 12.482mol

Xsolvent=(135.7kPa-13.3kPa)/135.7kPa=.902

Xsolvent=12.482/(12.482+x)=.902

re-arranged x=1.356 mol

therefor mass I2 = 344g