yes and yes.
Just look at the ionization. i is the number of particles.
NaCl ==> Na^+ + Cl^-
i = 2.
C6H12O6 ==> doesn't ionize. i = 1.
Na2SO4 ==> 2Na^+ + SO4^=
i = 2+1 = 3
Acetic acid.
HC2H3O2 ==> H^+ + C2H3O2^-
i = between 1 and 2.
The i done this way is theoretical and most solution are NOT ideal solutions. For example, acetic acid partially ionizes so i may be 1 if it is a very concentrated solution or closer to 1.5 or so in a more dilute solution and finally very close to 2 in VERY dilute solutions. In fact, this may be experimentally used to determine the percent ionization; i.e., determine how much from ideal behavior a solute is. But remember, too, that the solvent used in these detla T experiments need not be water and what may ionize in water may not ionize in camphor.
In the delta T formula how do you find i = ?
If its ionic like NaCl then i = 2
What if its like C6H12O6?
i = 1 ? or i = 3?
Since that is not ionic i would assume it will be 1
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Thank you! I feel about better. I'm at school right now and just got done with my chemistry test.
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