Posted by AJ on .
The freezing point depression of a 0.100 molality solution of NaCl(aq) solution is 0.34 deg C. Calculate the % dissociation of NaCl(aq) (Kf for water = 1.86 deg Cm^1)

Chemistry 
DrBob222,
Do you mean 0.34 degrees C. I don't see how it could be +0.34.

Chemistry 
AJ,
yeah .34 is the change in temp freezing

Chemistry 
DrBob222,
delta T = i*Kf*m
0.34 = i*1.86*0.1
solve for i.
Then i x 0.1 = effective molality which I will call em.
NaCl <==>Na^+ + Cl^=
Do an ICE chart for NaCl
NaCl will be 0.1y
Na^+ will be y
Cl^ will be y
The sum of NaCl, Na^+, and Cl^ will be equal to em
So you will have
0.1y + y + y = em
solve for y
Then (y/i)*100 = % ionization.
Post your work if you get stuck. 
Chemistry 
AJ,
i got an i of 1.83 then i divided it by 2 to get a dissociation of 91.5% is this correct?

Chemistry 
DrBob222,
The 1.83 for i is correct. But my instructions didn't say anything about dividing by 2. Follow my instructions calculate effective molality, then go through the ICE chart, then to y, then to percent ionization.