I am doing a latenitelab on Molecular Mass by Freezing Poing Depression. I am st
ID: 1048773 • Letter: I
Question
I am doing a latenitelab on Molecular Mass by Freezing Poing Depression. I am stuck on three questions.
1.Suppose you added 4.000 g of FP sample #1 instead of 2.000 g, what would happen to the freezing point temperature of the water?
2. Suppose you dissolve 154.286 g of sodium chloride in 2.00 L of water. What is the molality of the solution given that the molar mass of sodium chloride is 58.44 g/mol and the density of water is 1.000 g/mL?
3. Calculate the molar mass of FP sample 2.
Lab Notes:
1. I took an empty test tube and placed it on the balance. I zeroed the balance.
2. I put 10 mL of water into the tube. The weight was 10 g.
3. I took the water bath and set it to -15 degrees. I placed the test tube of water into the bath.
4. The water began to form ice at 0 degrees.
Experiment 2
1. I took an empty test tube and placed it on the balance.
2. I added 2 grams of FP sample 1.
3. The mass was 2 grams.
4. I added 10 mL of water.
5. The mass is now 12 g.
6. I placed the test tube in the bath.
7. Ice first began forming at -2.1.
8. I took another test tube and placed it on the balance.
9. I added 2 g of FP sample 2.
10. The mass is 2 g.
11. I added 10 mL of water.
12. Ice began to form on Sample 2 at -3.6.
We are given that the Van''t Hoff factor of Sample 1 is 1 and Sample 2 is 2.
Explanation / Answer
Freezing point experiment
1. If we took 4 g sample instead of 2 g, the resulting freezing point lowering would be greater than that observed for 2 g sample.
2. molality = grams of solute/molar mas x kg of solvent
as density of water = 1 g/ml,
volume of water = mass of water
molality of NaCl solution = 154.286 g/58.44 g/mol x 2 kg = 1.32 m
3. molar mass of sample = grams of sample/freezing point of sample x kg of solvent
For sample 2
molar mass = 2/-3.6 x 0.01 = 55.55 g/mol
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