Finding the Van't Hoft factor

Question

I am given the following values for a sulfuric acid aqueous soution.

Molal: .5

Mass: 6.9401

It lowered the freezing point to -4.88 *C

I need help solving this and would appreciate being walked through the steps.

Answer 1

Here is the full walkthrough, and then something important about the number you get.

Step 1 - the freezing point depression

Pure water freezes at 0.00 degrees C, so

delta Tf = 0.00 - (-4.88) = 4.88 degrees C

Depression is always taken as a positive quantity. If you carry the minus sign through you will get a negative i, which cannot happen.

Step 2 - the equation

delta Tf = i x Kf x m

where Kf for water is 1.86 degrees C/m and m is molality. Solve for i:

i = delta Tf / (Kf x m)

Step 3 - substitute

i = 4.88 / (1.86 x 0.5) = 4.88 / 0.93

i = 5.25

Now the part the question is really asking about

That answer is impossible, and saying so is the point of the exercise.

The van't Hoff factor is the number of particles one formula unit releases. For sulfuric acid the maximum conceivable value is 3:

H2SO4 gives H+ + HSO4-, then HSO4- gives H+ + SO4^2-, so at most 3 particles.

An experimental i of 5.25 exceeds that ceiling by a wide margin, so it is not telling you something new about sulfuric acid - it is telling you something is wrong with the measurement or the recorded data.

What the real answer should look like

The first proton of H2SO4 is fully dissociated in water, but the second is not: HSO4- is a weak acid with Ka2 about 1.2 x 10^-2, so at 0.5 molal only a modest fraction of it gives up its proton. The expected i is therefore between 2 and 3, and measured values at this concentration typically land near 2.2 to 2.5 - below 3 also because ion pairing in a concentrated electrolyte makes some ions behave as one particle.

So the conclusion your write-up wants is: sulfuric acid dissociates completely in its first step and only partially in its second, giving an i meaningfully greater than 1 but less than 3.

Where 5.25 probably came from

Work backwards. To get a legitimate i of about 2.4 from a 4.88 degree depression you would need m = 4.88 / (1.86 x 2.4) = 1.09 molal, roughly twice the recorded 0.5. That points at the molality rather than the thermometer.

Check it against your other number: 6.9401 g of H2SO4 is 6.9401 / 98.08 = 0.0708 mol. For that to be 0.5 molal you need 0.0708 / 0.5 = 0.142 kg of water. If the actual solvent mass was closer to 65 g, the molality is about 1.09 and everything reconciles. Re-weigh or re-read the solvent mass and the whole problem falls into place.

Supercooling is the other usual suspect, since it makes the observed freezing point read too low and inflates i - but it rarely accounts for a factor of two.