How Much Iodine to Dissolve in Chloroform to Lower Vapour Pressure by 13.3 kPa?

[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 can't do this question, i think in involves raults law, P=XP_A
anyhelp would be good. thanks



p.s. answer is 1.32 x 101 g

 

[Borek]

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?

 

[geffman1]

thnks for the reply, umm 122.4kpa

 

[bluesman4509]

Hey,

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

P_solution=x_solvent.P_solvent

so P_solution=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 don't know where went wrong.

 

[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, I've tried but get lost. thanks

 

[bluesman4509]

Ok,

so X_solvent=mol_solvent/(mol_solute+mol_solvent), we are looking to find mol_solute (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

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

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

re-arranged x=1.356 mol

therefor mass I₂ = 344g