Calculating fractions of moles of solute that dissociated from the observed van't hoff factor.?

Question

I thought that the van't hoff factor is the moles of particles in solution divided by moles of solute. Because the observed van't hoff factor is less than two for my sodium chloride solution, wouldn't the equation be (2X+Y)/(X+Y)=i ? Where x is the moles of solution that will be dissociated, and y is the moles of solute that will remain clumped. assuming X+Y=1, then 2X+Y=i =>
2X+1-X=i => X=1-i ...and the percent of dissociated sodium chloride would be X/1 multiplied by 100?
Can someone tell me why this is wrong?

Answer 1

Your equation is completely wrong.

i is a correction factor for the fact that you have more particles than you would expect from the molecular formula.

If you have N particles of a weak electrolyte that has degree of dissociation a and in theory gives n ions for 100% dissociation according to the molecular formula, then the particles you have is the number of ions that come from the dissociation plus the non dissociated particles.

The free ions are n*a*N
The non dissociated are (1-a)N
Thus the total number of particles in solution is
naN+(1-a)N= (na+(1-a))N

The ideal would be when you have 0 dissociation, thus N particles

so i=(na+(1-a))N/N=na+(1-a) => a=(i-1)/(n-1)

NaCl is a strong electrolyte but at high concentrations you have ion pairing and i<2, so you could treat it as a weak electrolyte and determine a. In your case n=2, i you have an experimental value, so you just plug the values in the equation.