1. Dismiss Notice
  2. Hey Guest! Check out the 3 MCAT Study Plan Options listed in the 'stickies' area at the top of the forums (BoomBoom, SN2ed, and MCATJelly). Let us know which you like best.

    Also, we now offer a MCAT Test-Prep Exhibitions Forum where you can ask questions directly from the test-prep services.
    Dismiss Notice
  3. This forum is for support and discussion only. Please promote test prep materials/services (including AMAs) in the Special Offers subforum only. Thanks!
    Dismiss Notice
Dismiss Notice

Interview Feedback: Visit Interview Feedback to view and submit interview information.

Interviewing Masterclass: Free masterclass on interviewing from SDN and Medical College of Georgia

analytical chemistry ?

Discussion in 'MCAT: Medical College Admissions Test' started by Mr. Z, Aug 13, 2002.

  1. Mr. Z

    Mr. Z Senior Member
    7+ Year Member

    Joined:
    Apr 19, 2002
    Messages:
    451
    Likes Received:
    0
    Try these on for size...


    1. In order for two compounds to be separated by NaHCO3 extraction, they must have:

    a) different dipole moments
    b) different acid base properties
    c) different boiling points
    d) different molecular weights


    2. Which pair of compounds could be separated most readily by solvent extraction.

    a) aminocycylohexane and chlorocyclohexane
    b) cyclohexanol and cyclohexanone
    c) phenol and p-cresol
    d) butanoic acid and 2 methylpropanoic acid

    3. Carboxylic acids are generally separated as their methyl esters in gas chromatography. Why are the carboxylic acids themselves difficult to separate by this technique?

    a) carboxylic acids all have the same nominal polarity
    b) carboxylic acids are insoluble in most organic solvents while
    their methyl esters are soluble.
    c) carboxylic acids decompose readily in a gas chromatographic
    column
    d) carboxylic acids do not vaporize readily but their methyl
    esters do so easily
     
  2. freakazoid

    freakazoid Guy Friend Extraordinaire
    7+ Year Member

    Joined:
    Jun 16, 2002
    Messages:
    217
    Likes Received:
    0
    1)B
    2)B
    3)D

    The last one I wasn't sure about, but I believe gas chromatography works by separating compounds of different volatilities . . .
     
  3. rCubed

    rCubed taiko master
    10+ Year Member

    Joined:
    Jul 14, 2002
    Messages:
    416
    Likes Received:
    0
    hmm..i thought that #1 was a.
    since NaHCO3 is a salt, it comprises the aqueous layer, and for a compound to be soluble in it, it must contain a dipole moment, otherwise it would be soluble in the organic layer
     
  4. Mr. Z

    Mr. Z Senior Member
    7+ Year Member

    Joined:
    Apr 19, 2002
    Messages:
    451
    Likes Received:
    0
    freakazoid is on the money, nice job.
     
  5. limit

    limit Molesting my inner-child
    10+ Year Member

    Joined:
    Jun 21, 2000
    Messages:
    570
    Likes Received:
    1
    Close. NaHCO3 is used in extraction to deprotonate strong acids in order to make them more polar, hence allowing them to dissolved and be removed with the aqueous layer. This makes the answer B (acid base characteristics)
     
  6. rCubed

    rCubed taiko master
    10+ Year Member

    Joined:
    Jul 14, 2002
    Messages:
    416
    Likes Received:
    0
    thanks limit, i appreciate the clarification

    seperation is by far the most difficult subject for me on this exam, followed by....this is sad: translational motion
     

Share This Page