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

AI Thread Summary
The discussion focuses on calculating the mass of iodine (I2) needed to dissolve in chloroform (CHCl3) to reduce its vapor pressure by 13.3 kPa from an initial pressure of 135.7 kPa. Participants reference Raoult's Law to determine the new pressure of the solution and the mole fractions involved. The calculations reveal that approximately 344 grams of iodine is required to achieve the desired vapor pressure reduction. There is some confusion regarding the initial calculations and the values used, but consensus is reached on the correct approach. The final conclusion emphasizes the importance of accurately applying Raoult's Law in such calculations.
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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=XPA
anyhelp would be good. thanks



p.s. answer is 1.32 x 101 g
 
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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?
 
thnks for the reply, umm 122.4kpa
 
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 don't know where went wrong.
 
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
 
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
 
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