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)
Do you mean -0.34 degrees C. I don't see how it could be +0.34.
yeah -.34 is the change in temp freezing
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 0.1-y
Na^+ will be y
Cl^- will be y
The sum of NaCl, Na^+, and Cl^- will be equal to em
So you will have
0.1-y + y + y = em
solve for y
Then (y/i)*100 = % ionization.
Post your work if you get stuck.
i got an i of 1.83 then i divided it by 2 to get a dissociation of 91.5% is this correct?
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.
To calculate the percent dissociation of NaCl(aq), we need to first determine the amount of dissociated ions in the solution. The freezing point depression is a colligative property that depends on the number of solute particles present in the solution.
The freezing point depression (∆Tf) is given by the formula:
∆Tf = Kf * m * i
Where:
∆Tf = Freezing point depression
Kf = Freezing point depression constant for the solvent
m = Molality of the solution (moles of solute per kilogram of solvent)
i = Van't Hoff factor (number of particles into which the solute dissociates)
In this case, the solute is NaCl, which dissociates into one Na+ ion and one Cl- ion. Therefore, the Van't Hoff factor (i) for NaCl is 2.
Given that ∆Tf = 0.34°C, Kf for water = 1.86°Cm^-1, and the molality (m) of the solution is 0.100 molal, we can rearrange the formula to solve for the percent dissociation:
∆Tf = Kf * m * i
0.34 = 1.86 * 0.100 * 2
Now, let's solve for the percent dissociation.
First, calculate the number of moles of NaCl in the solution:
moles of NaCl = molality (m) * mass of solvent (in kg)
moles of NaCl = 0.100 * (mass of solvent in kg)
Assuming we have 1 kilogram (kg) of water, the mass of solvent is equal to 1 kg.
moles of NaCl = 0.100 * 1
moles of NaCl = 0.100
Since NaCl dissociates into one Na+ and one Cl- ion, the total number of dissociated particles is equal to the number of moles of NaCl.
Number of dissolved particles = moles of NaCl = 0.100
Now, we can calculate the percent dissociation:
% dissociation = (Number of dissociated particles / Total number of particles) * 100
% dissociation = (0.100 / 0.100) * 100
% dissociation = 100
Therefore, the percent dissociation of NaCl(aq) is 100%.