lian LAB ACTIVITY 14.1 Pre-Lab Questions Date 1 Use your textbook to find the re
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lian LAB ACTIVITY 14.1 Pre-Lab Questions Date 1 Use your textbook to find the reported value for the freezing point depression constant for water value in Table 14.3 on the data sheet, page 459. You will need this constant for your calculations. 2 Calculate the freezing point depression observed in the following solutions, showing your work in the spaces provided: a 0.75 mol of ethylene glycol (C,H,0) dissolved in 50.0 g of water b 0.75 mol of calcium chloride (CaCl,) dissolved in 50.0 g of water. Calculate the molar masses of sodium chloride, glucose, and urea. Record them in Table 14.1 on the next page and in Table 14.4 on p. 459 in the data sheet. 3 4 The solutions you make in this activity will all contain the same mass of solute in the same mass of water. Predict the relative size of the freezing point depression you will observe due to each solute. Use"1" for the lowest FP depression, "3* for the highest, and -2 for the intermediate FP depression. Record your predictions in Table 14.1. Explain how you arrived at your rankingExplanation / Answer
Ans. #2a. Molality of solution = Moles of solute / Mass of solution in kg
= 0.75 mol / 0.050 kg
= 15.0 m
Depression in freezing point of the solution is given by-
dTf = i Kf m - equation 1
where, i = Van’t Hoff factor. [ i = 1 for non-electrolyte]
Kf = molal freezing point depression constant of solvent = 1.860C / m
m = molality of the solution
dTf = Freezing point of pure solvent – Freezing point of solution
= Freezing point depression
Putting the values in equation 1-
dTf = 1 x (1.86 0C m-1) x 15.0 m
Hence, dTf = 27.90C
#2b. Putting the values in equation 1-
dTf = 3 x (1.86 0C m-1) x 15.0 m = 83.70C
#3. Molar mass of NaCl = 58.442468 g/ mol
MW, glucose C6H12O6 = 180.15768 g/ mol
MW, Urea CO(NH2)2 = 60.05564 g/ mol
#4. Let the mass of solutes taken = X gram. And, mass of solvent taken is 1.0 kg.
# NaCl: Moles of NaCl = X g / (58.442468 g/ mol) = 0.0171X mol
Molality of NaCl = 0.0171X mol / 1.0 kg = 0.0171X m
Now,
dTf, NaCl = 2 x (1.86 0C m-1) x 0.0171X m = 0.063612X 0C
# Glucose: Moles of Glucose = X g / (180.15768 g/ mol) = 0.00555X mol
Molality of glucose = 0.00555X mol / 1.0 kg = 0.00555X m
Now,
dTf, glucose = 1 x (1.86 0C m-1) x 0.00555X m = 0.010323X 0C
# Urea: Moles of Urea = X g / (60.05564 g/ mol) = 0.01665X mol
Molality of urea = 0.01665X mol / 1.0 kg = 0.01665X m
Now,
dTf, urea = 1 x (1.86 0C m-1) x 0.01665X m = 0.030969X 0C
# Order of freezing point depression = NaCl < Urea < Glucose
#5. Step 1: Let the molar mass of unknown = X g/mol
Now,
Moles of compound = Mass / Molar mass
= 0.461 g / (X g/mol) = (0.461 / X) mol
Molality of solution = (0.461 / X) mol / 0.010 kg = (46.1/X) m
# Depression in freezing point, dTf =
Freezing point of pure solvent – Freezing point of solution
= 0.00C – (-2.30C) = 2.30C
# Step 2: Depression in freezing point of the solution is given by-
dTf = i Kf m - equation 1
where, i = Van’t Hoff factor. [ i = 1, assumed]
Kf = molal freezing point depression constant of solvent = 1.860C / m
m = molality of the solution
dTf = Freezing point of pure solvent – Freezing point of solution
Putting the values in equation 1-
2.30C = 1 x (1.86 0C m-1) x (46.1 / X) m
Or, X = (1.860C x 46.1) / 2.30C
Hence, X = 37.28
Therefore, molar mass of sugar = X g/ mol = 37.28 g/ mol
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