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Author Topic: Properties of Solutions  (Read 1482 times)
holtb
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« on: May 20, 2007, 01:41:05 PM »

I don't have any problem with the math part such as molarity, molality, and so on.  However, when it comes to determining what dissolves best in water or another substance I'm completely at a loss.  For instance a question like:

Which of the following in each pair is more likely to be soluble in hexane, C6H14? (i) CH3CH2CH2CH3 or CH3CH2OH (ii) CCl4 or CaCl2 (iii) C6H6 or C6H5OH (phenol)

OR

Which of the following pairs is most likely to be heterogeneous?
        hexane, CH3(CH2)4CH3; and octane, CH3(CH2)6CH3
        methanol, CH3OH; and water, H2O
        acetic acid, CH3COOH; and water, H2O
        iodine, I2; and benzene, C6H6
        potassium chloride, KCl; and carbon tetrachloride, CCl4

These are off our first homework assignment for an online summer class so I don't necissarily need them answered, just clarified as to how I know or what I need to do to figure it out.  I'm sorry if this is to broad or what not.
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« Reply #1 on: May 20, 2007, 03:00:25 PM »

Ok what you want to look at here is the intermolecular forces between each substance. Basically is the compound polar or not.

Remember the age old saying water and do not mix or water and gasoline do not mix. That is because water is polar and gasoline is not.

Like dissolves like.  All the water molecules clumped together and leaves all the gasoline out.

so compare the polarities of each pair of molecules. For instance, hexane is a nonpolar alkane

(i) CH3CH2CH2CH3 or CH3CH2OH

here you have butane and very similar nonpolar alkane and ethanol which is has a polar group on the end. butane is most "like" hexane so it will dissolve in it.

by the way, how did you hear about us?



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holtb
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« Reply #2 on: May 21, 2007, 04:42:47 PM »

I just came across the site when I was searching for something that could possibly help explain some chem stuff to me.
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« Reply #3 on: May 21, 2007, 04:44:16 PM »

cool please tell your friends if you are in a class
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holtb
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« Reply #4 on: May 21, 2007, 04:49:22 PM »

I will definately do that...Thanks for the help!
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