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

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SUMMARY

The discussion focuses on calculating the mass of iodine (I2) required to dissolve in 1 liter of chloroform (CHCl3) to reduce its vapor pressure by 13.3 kPa. The initial vapor pressure of chloroform at 70.0°C is 135.7 kPa. Using Raoult's Law, the final calculation reveals that 344 grams of iodine must be added to achieve the desired pressure reduction. The participants confirmed the calculations and clarified the necessary steps to derive the solution accurately.

PREREQUISITES
  • Understanding of Raoult's Law and its application in vapor pressure calculations
  • Familiarity with molar mass calculations, specifically for iodine (I2) and chloroform (CHCl3)
  • Basic knowledge of solution chemistry and mole fraction concepts
  • Ability to perform unit conversions, particularly between grams and moles
NEXT STEPS
  • Study Raoult's Law and its implications in colligative properties
  • Learn how to calculate mole fractions in solutions
  • Explore the concept of vapor pressure lowering in various solvents
  • Investigate the effects of temperature on vapor pressure and solubility
USEFUL FOR

Chemistry students, chemical engineers, and professionals involved in solution chemistry and vapor pressure calculations will benefit from this discussion.

geffman1
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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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