Question

If 100 percent of the NaCL molecules dissociated, the Van't Hoff Factor, I, would be equal to 2 for NaCI. We assumed I=2 when we calculated the K_f of the NaCl solution.

a Now calculate the opposite. Experimentally determine the Van't Hoff factor ( I ) for each NaCl trial, using kf=1.86 C/m and the calculated molality of your NaCl solution and the change in temperature for that trial. Calculate the average "I" value.

i = (deltaT_f) / (k_f x m)

b How did making the assumption. that NaCl completely dissociated (as we did) affect your calculation of the freezing point depression constant K_f?

c which solute produced a more accurate k_f value? Report the results of both and compare.

Answer 1

Hi,

Hope you are doing well.

(a)

I will give you a sample experiment trial that I have done earlier. The datas were as shown below.

Mass of NaCl in solution=31.65g

No.of moles of NaCl is,

Amount of water used to make NaCl solution,

Molarity of NaCl solution is given by,

In this experiment, the value of depression of freezing point was measured as,

From this, We can calculate average value of Van't Hoff Factor, i as,

​​​

We have,

(b)

From this experiment, We can say that NaCl can't undergo complete dissociation in normal condition. If NaCl is completely ionised, then we would have got a value of 2 in our experiment. If we assume that i=2, then we will get wrong answer for ,

(c)

Solutes which can't undergo dissociation can give more accurate values for K_f since the i value makes no significance in this case.

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