The solutes used in this experiment were non-electrolytes. Let\'s consider an ex

Question

The solutes used in this experiment were non-electrolytes. Let's consider an experiment where a student compares the freezing point depression of a non-electrolyte solute to that of an ionic solute. The student adds 0.25 moles of table sugar, sucrose, to test tube #1 and 0.25 moles of an ionic solute, NaCl, to test tube #2. The student measures the freezing point of both solutions and determines that in tube # 1,the change in temperature is 3.0 C, while in tube # 2, the change in temperature is 5.7 C. Suggest an explanation for this observation.

Explanation / Answer

Answer:

we know that

depression in freezing point is given by

dTf = i x kf x m

given

0.25 moles of non ionic solute

and

0.25 moles of NaCl

assuming equal amount of solvent

molalities of both solutions are same

also

Kf is a constant

so

dTf depends on i

now

i is called vanthoff factor ,

it is the number of particles in the solution

now

for a non ionic solution

it does not dissociate in the solution

it stays as a big single particle

so

i = 1 for non ionic solutions

now

consider NaCl

NaCl ---> Na+ + Cl-

we can see that

there are two particles

so

i = 2 for NaCl

now

dTF non ioninc / dTf NaCl = i for non ionic / i for NaCl

dTf non ionic / dTf NaCl = 1 /2

dT NaCl = 2 x dTf non ionic

as we can see

dT for NaCl is 5.7 and dT for non ionic is 3

note : it is not exactly two times

becauses

NaCl may not dissociate 100%

so

the i value of NaCl will be a little less than 2

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