Experiment questions:
Experiment 1 results
mass of beaker:116.944 g
mass of dodecanoic:11.629 g
hot water: 175 C
warm water: 10 C
Average Fressing pt of Dodecanoic = 42.7
Experiment 2 results:
Unknown C solution mass: 2.091g
Hot water: 175C
warm water: 10C
average FP of UK C 71.8C
From Part 2, take the average of the two freezing point experiments to get the freezing point for the dodecanoic acid solution and determine T, the magnitude of the freezing-point depression (a positive number) for the solution of the unknown solute and the dodecanoic acid.
Applying the formula and dividing, (the van't Hoff factor I is equal to 1 for
these compound because they are nonelectrolytes).
Jessica, I don't understand the problem.
That's the same issue i'm having. This is directly from my printed lab sheet.
To find the freezing point depression (ΔT) for the solution of the unknown solute and dodecanoic acid, you need to calculate the average freezing point (FP) of dodecanoic acid solution and then subtract it from the freezing point of the unknown solute.
First, calculate the average freezing point (FP) of dodecanoic acid solution:
Average FP = (FP Experiment 1 + FP Experiment 2) / 2
Average FP = (42.7 + 71.8) / 2
Average FP = 114.5 / 2
Average FP = 57.25°C
Next, subtract the average freezing point of dodecanoic acid solution from the freezing point of the unknown solute to find the freezing point depression (ΔT):
ΔT = FP Unknown solute - Average FP of dodecanoic acid solution
ΔT = 71.8 - 57.25
ΔT = 14.55°C
Therefore, the magnitude of the freezing-point depression (ΔT) for the solution of the unknown solute and dodecanoic acid is 14.55°C.