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
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.
Thank you! I feel about better. I'm at school right now and just got done with my chemistry test.
In the delta T formula, the "i" represents the van't Hoff factor, which is the number of particles that a compound dissociates into when it dissolves in a solvent.
For ionic compounds like NaCl, where the compound dissociates into ions, the van't Hoff factor (i) is equal to the number of ions produced. In the case of NaCl, it dissociates into Na+ and Cl- ions, so the van't Hoff factor is 2.
However, for non-ionic compounds like C6H12O6 (glucose), the molecules do not dissociate into ions when they dissolve. Therefore, the van't Hoff factor for non-ionic compounds is generally considered as 1.
So, in the case of C6H12O6, the van't Hoff factor would be 1.