Molar Mass by Freezing Point Depression Short Answer Questions From my lab I col

Question

Molar Mass by Freezing Point Depression Short Answer Questions

From my lab I collected the following data.

Experiment 1:

mass of water = 10.000g
freezing point of the solvent = 0C

Experiment 2:

mass of sample FP1 added = 2.000g
total mass of water + FP1 sample = 12.000g
freezing point of the FP1 solution = -2.1C

mass of sample FP2 added = 2.000g
total mass of water + FP2 sample = 12.000g
freezing point of the FP2 solution = -3.8C

Molar Mass by Freezing Point Depression SHORT ANSWER Experiment 1: Measure the Freezing Point of Pure Water Lab Results 1. What was the mass of water used in this experiment? Data Analysis 2. How does the freezing point measured for water compared to what you expected? Experiment 2: Measure the Freezing Point of a Solution of an Unknown Substance Lab Results . At what temperature did the solution of FP sample 2 freeze? Lab Results 2. Calculate the change in temperature for the solution of FP sample 1 3. Calculate the molar mass of FP sample 2. Conclusions . In winter, you can buy a variety of salts to use on your sidewalk. Two of the most popular choices are NaCl and CaCl2 Which of the two salts is more effective? Justify your answer 2. Suppose a student performed a similar experiment with a different ionic unknown. Given the data in the table below, what is the van 't Hoff factor of the unknown substance? Kf of water freezing point of pure water 0.0 °C molality of solution freezing point of solution 1.86 °C/m 0.512 mol/kg 2.9 °C

Explanation / Answer

Freezing point experiment

Experiment 1

1. Mass of water used = 10 g

2. Freezing point of pure solvent (water) is 0 oC which is expected.

Experiment 2

1. Freezing point of solution = 3.8 oC

2. Change in freezing point = 0 - (-3.8) = 3.8 oC

3. molality of solution FP2 = 3.8/1.86 = 2.04 m

molar mass of FP2 = 2/2.04 x 0.01 = 98.04 g/mol

4. Most likely the unknown is sodium oxalate

Conclusions

1. In winter CaCl2 with van't hoff factor of 2 is more effective as more ions are present which produces more heat as compared to NaCl with 2 ions only and thus are preferred choice for winter.

2. van't hoff factor = 2.9/0.512 x 1.86 = 3

3.

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