August 06, 2020, 04:02:14 AM
Forum Rules: Read This Before Posting


Topic: Vapor Pressure question  (Read 4849 times)

0 Members and 1 Guest are viewing this topic.

Offline 21385

  • Regular Member
  • ***
  • Posts: 55
  • Mole Snacks: +2/-1
Vapor Pressure question
« on: October 07, 2007, 04:24:40 AM »
A solution of LiCl in a mixture of water and methanol (CH3OH) has a vapor pressure of 39.4 mmHg at 17C and 68.2 mmHg at 27C. The vapor pressure of pure water is 14.5 mmHg at 17C and 26.8 mmHg at 27C, and the vapor pressure of pure methanol is 82.5 mmHg at 17C and 140.3 mmHg at 27C. What is the composition of the solution (in mass percent)?

I have attempted at solving this problem but my solution and answers were different from the textbook's answer.

Here is the answer from the textbook (Rightclick and click on view image to see the image better) Also, the symbol "_" is a minus sign.



But I have a problem with the textbook solution. It does not consider Raoult's Law, which says that the solution's vapor pressure will be less because of the dissolved LiCl.

Here's my solution


I don't know why the answers and solutions are different. Can anyone help me? Thanks a lot.

Offline 21385

  • Regular Member
  • ***
  • Posts: 55
  • Mole Snacks: +2/-1
Re: Vapor Pressure question
« Reply #1 on: October 08, 2007, 11:10:45 AM »
I really need help for this. Can anyone tell me where I went wrong or how would they approach this problem?

Offline Yggdrasil

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 3211
  • Mole Snacks: +482/-21
  • Gender: Male
  • Physical Biochemist
Re: Vapor Pressure question
« Reply #2 on: October 08, 2007, 12:02:23 PM »
The book's solution does take into account Raoult's law.  Because LiCl is dissolved in the solution, Xwater + Xmethanol < 1.  So, the LiCl lowers the vapor pressure of the solution by lowering Xwater and Xmethanol.

Your equations are wrong because you treat the (1/(1+y)) factor incorrectly.  It should be (1/(1+XLiCl)), but this just makes the equation unnecessarily difficult to work with.

Sponsored Links