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

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Discussion Overview

The discussion revolves around calculating the mass of iodine (I2) required to dissolve in chloroform (CHCl3) to achieve a specific reduction in vapor pressure, specifically by 13.3 kPa. Participants explore the application of Raoult's Law in this context, addressing the theoretical and mathematical aspects of the problem.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant suggests that the problem involves Raoult's Law, specifically the equation P = XPA, and seeks assistance in solving it.
  • Another participant questions whether the vapor pressure of iodine should be considered in the calculations.
  • A participant provides a calculation leading to a vapor pressure of 122.4 kPa after the iodine is added.
  • One participant presents a detailed solution, arriving at a different mass of iodine (344.23 grams) than previously suggested, indicating a discrepancy in the calculations.
  • Another participant confirms the correctness of the previous answer after consulting with a teacher and requests clarification on the numbers used in the calculations.
  • A later reply outlines the steps to find the mole fraction of the solvent and the mass of iodine needed, arriving again at 344 grams.

Areas of Agreement / Disagreement

There is no consensus on the correct mass of iodine required, as participants present different calculations and results. The discussion remains unresolved regarding the correct approach and final answer.

Contextual Notes

Participants express uncertainty about whether to include the vapor pressure of iodine in their calculations, which may affect the results. Additionally, there are unresolved steps in the mathematical reasoning presented.

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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