Dissociated electrolyte?

Question

calculate the freezing point in degrees C of a 0.100 molality (not molarity) aqueous solution of a weak electrolyte that is 7.5% dissociated.

please, i dont want the answer, i don't know how to start the problem with the percentage and molality.
thanks

Answer 1

We can't help you unless you provide the number of ions in the molecular formula. E.g. is it of the type AB, A2B, AB2, A3B2 or what?

The freezing point of the aqueous solution will be Tf= Tf0-ΔTf where Tf0 is the freezing point of the solvent. For water Tf0=0 deg Celsius thus Tf = -ΔTf

but ΔTf=i*Kf*m

where m is the molality of the electrolyte and i the van't Hoff coefficient. This coefficient is introduced in order to correct for the fact that in electrolyte solution you will have more particles than you would expect according to the molecular formula. So the value of i should represent the number of particles that are truly in solution.

For strong electrolytes i =n where n is the number or ions produced by the dissociation according to the molecular formula e.g. for NaCl i=2 since it dissociates into 1 Na+ and 1 Cl-, for CaCl2 i=3 (dissociates into 1 Ca(+2) and 2 Cl-) and so on.

For weak electrolytes it is easily derived that i=(1-a) +n*a where n is the same as above and a is the degree of dissociation. In your case a=7.5% =0.075
You need to know n, in order to proceed. If we assume for simplicity that the formula is AB then n=2 and
i =1-0.075+2*0.075 =1.075

So Tf= -ΔTf= -i*Kf*m = -1.075*1.858*0.1 = -0.1997 =-0.20 deg C

**** Deriving the equation i=1-a+na
.. .. .. .. .. .. AxBy <=>xA(+y) + yB(-x) and x+y=n
Initial .. .. .. .. m
Dissociate.. .. z
Produce .. .. .. .. .. .. .. xz .. .. .. .. yz
At equil .. .. m-z .. .. .. .xz .. .. .. . ..yz

So at equil the total molality of all dissolved particles is
m(total)=[AxBx] + [A(+y)]+ [B(-x)] = m-z+xz+yz = m-z+(x+y)z
but by definition a=z/m=> z=am and also we said that x+y=n. So
m(total)= m-am+nam = (1-a+na)*m
ΔTf= Kf*m(total) and also
ΔTf = i*Kf*m
Thus Kf*m(total)=i*Kf*m =>
m(total)=i*m =>
(1-a+na)*m=i*m =>
i=1-a+na