# percent dissociation

Would these answers be correct? . Of 100% of the NaCL molecules dissociated, the Van't Hoff factor i, would be equal to 2 for NaCl. We assumed it equaled two when we figured out Kb of the NaCl solution. a. Now we want it find the percent dissociation of the NaCl molecules by finding i using the equation below (use Kb = 0.512℃/m and your calculated molarity of NaCl solution ) show all work i = ( ▵Tb) / Kb x m) i experimental = 1.8 [ {1.8/2} * 100% = 90% dissociation] i theoretical = 2 (100% dissociation)

my answer - % dissociation = (1.74/2) *100% = 86.9%

b. . Do you feel comfortable in making the assumption that the NaCl in your experiment was completely dissociated? Why or why not? I do not feel comfortable in making the assumption that that the NaCl in my experiment was completely dissociate because it lead to a 16.6% difference in my experiment

Thank you

So when $\ce{NaCl}$ is dissolved in water all the sodium and chloride ions are truly dissociated. There are no "molecules" of $\ce{NaCl}$ floating around in the aqueous solution.
Imagine the solution this way. Instead of the water molecules bouncing around in the solution more or less randomly, the positive charged $\ce{Na^+}$ ion attracts the more negative oxygen atom of the water molecule leaving the positive hydrogen atoms on the outside of this first "shell" of ligands. Now a second "shell" of water molecules will be attracted to the first (Oxygen to hydrogens) and so on. Now each shell has more and more water molecules but the binding energy decreases as the shell number increases. So the gist is that there is an exponential decrease in the ordering of the water molecules. Ultimately the ordering becomes random again as in the bulk of the water.