### Freezing Point DepressionProblems #1-10

Problem #1: How many grams of pyrazine (C4H4N2) would have to be dissolved in 1.50 kg of carbon tetrachloride to lower the freezing point by 4.4 °C? The freezing point constant for carbon tetrachloride is 30. °C/m.

Solution:

Δt = i Kf m

4.4 °C = (1) (30. °C kg mol¯1) (x / 1.50 kg)

4.4 °C = (1) (20. °C mol¯1) (x)

x = 0.22 mol

0.22 mol times 80.0896 g/mol = 17.6 g (I'll ignore sig figs and leave it at three. I'm such a rebel!)

Problem #2: When 0.258 g of a molecular compound, benzoic acid, was dissolved in 40.0 g of benzene, the freezing point of the solution was lowered to 5.23 °C. What is the molecular weight of the benzoic acid?

Solution:

We look up the Kf for benzene, finding it to be 5.12 °C/m. The freezing point of benzene is found to be 5.5 °C.

Δt = i Kf m

0.27 °C = (1) (5.12 °C kg mol¯1) (x / 0.0400 kg)

0.27 °C = (128 °C mol¯1) (x)

x = 0.00211 mol

0.258 g / 0.00211 mol = 122 g/mol (to three sig figs)

Problem #3: When 92.0 g of a molecular compound was dissolved in 1000. g of water, the freezing point of the solution was lowered to −3.72 °C. What is the molecular weight of the compound?

Solution:

Δt = i Kf m

3.72 °C = (1) (1.86 °C kg mol¯1) (x / 1.000 kg)

3.72 °C = (1.86 °C mol¯1) (x)

x = 2.00 mol

92.0 g /2.00 mol = 46.0 g/mol (to three sig figs)

Problem #4: What is the freezing point depression when 62.2 g of toluene (C7H8) is dissolved in 481 g of naphthalene? The freezing point constant for naphthalene is 7.00 °C/m.

Solution:

62.2 g / 92.1402 g/mol = 0.675058 mol Δt = i Kf m

x = (1) (7.00 °C kg mol¯1) (0.675058 mol / 0.481 kg)

x = 9.82 °C <--- that's not the new freezing point, that's the amount the freezing point is depressed

Problem #5: How many grams of pyrazole (C3H4N2) would have to be dissolved in 736 g of camphor to lower the freezing point by 15.0 °C? The freezing point constant for camphor is 40. °C/m.

Solution:

Δt = i Kf m

15.0 °C = (1) (40. °C kg mol¯1) (x / 0.736 kg)

15.0 °C = (54.3478 °C mol¯1) (x)

x = 0.276 mol

68.0786 g/mol times 0.276 mol = 18.8 g (to three sig figs)

Problem #6: What is the freezing point of a solution prepared by adding 140. g trichothecin (C19H24O5) to 0.746 kg of benzene? The freezing point of pure benzene is 5.5 °C. The freezing point constant for benzene is 5.12 °C/m.

Solution:

140. g / 332.39 g/mol = 0.421192 mol

Δt = i Kf m

x = (1) (5.12 °C kg mol¯1) (0.421192 mol / 0.746 kg)

x = 2.89 °C <--- this is the amount of freezing point depression, not the freezing point

5.5 °C − 2.89 °C = 2.6 °C (to two sig figs this is the new freezing point)

Problem #7: What is the freezing point depression when 309 g of isoprene (C5H8) is dissolved in 747 g of ethyl ether? The freezing point constant for ethyl ether is 1.79 °C/m.

Solution:

309. g / 68.1182 g/mol = 4.536233 mol

Δt = i Kf m

x = (1) (1.79 °C kg mol¯1) (4.536233 mol / 0.747 kg)

x = 10.87 °C

Problem #8: What is the freezing point of a solution prepared by adding 239.0 g of copper(II) sulfate pentahydrate to 4.00 liters of water? The freezing point depression of water is 1.86 °C/m.

Solution:

239.0 g / 249.681 g/mol = 0.95722 mol Δt = i Kf m

x = (2) (1.86 °C kg mol¯1) (0.95722 mol / 4.00 kg)

x = 0.89 °C

The solution freezes at −0.89 °C.

Note the use of a van 't Hoff factor of 2 for CuSO4.

Problem #9: A solution that contain 55.0 g of ascorbic acid (vitamin C) in 250. g of water freezes at −2.34 °C. Calculate the molar mass (in g/mol) of the solute.

Solution:

1) Use ΔT = i Kf m

2.34 = (1) (1.86) (x)

x = 1.258 mole/kg

Note: the van 't Hoff factor for ascorbic acid is 1 since it does not ionize in solution (it is a weak acid, so it actually does ionize a tiny bit, but we are ignoring it for this problem)

2) molecular weight, method #1

1.258 mol is to 1 kg as x is to 0.250 kg

x = 0.3145 mol

55.0 g / 0.3145 mol = 175 g/mol

3) molecular weight, method #2

55 g/0.250 kg = 220 g/1 kg

220 g is to x as 1.258 molal is to 1 molal

x = 175 g

175 g is the mass of ascorbic acid in the 1 molal solution, which is 1 mole of solute per 1 kg of solvent.

Conclusion: molar mass is 175 g/mol

Problem #10: When 1.150 grams of an unknown non-electrolyte dissolves in 10.0 grams of water, the solution freezes at −2.16 °C. What is the molecular weight of the unknown compound? Kf for water = 1.86 °C/ m.

Solution:

1) Determine how many moles of the compound dissolved:

Δt = i Kf m

2.16 °C = (1) (1.86 °C kg mol¯1) (x / 0.0100 kg)

2.16 °C = (186 °C mol¯1) (x)

x = 0.0116129 mol

2) Determine molecular weight:

1.150 g / 0.0116129 mol = 99.0 g/mol

Bonus Problem #1: Acetic acid (CH3COOH) is a polar molecule and can form hydrogen bonds with water molecules. Therefore, it has a high solubility in water. Yet acetic acid is also soluble in benzene (C6H6), a nonpolar solvent that lacks the ability to form hydrogen bonds. A solution of 3.80 g of CH3COOH in 80.0 g C6H6 has a freezing point of 3.46 °C. What is the molar mass of the solute?

Solution:

1) Some looking up yields the following values for benzene:

freezing point = 5.49 °C
cryoscopic constant (Kf = 5.12 °C kg mol¯1

2a) We need to determine how many moles of acetic acid were dissolved:

Δt = i Kf m

2.03 °C = (1) (5.12 °C kg mol¯1) (x / 0.0800 kg)

2.03 °C = (64 °C mol¯1) (x)

x = 0.031719 mol

Note the use of 1 for the van 't Hoff factor. This is done because acetic acid is dissolving in a nonpolar solvent. Typically, ions do not form in a nonpolar solvent.

2b) Here is an alternate pathway to the moles:

Δt = Kf m <--- setting i = 1

m = Δt / Kf

m = 2.03 °C / 5.12 °C kg mol¯1 = 0.396484 mol/kg

moles ---> (0.396484 mol / kg) (0.0800 kg) = 0.031719 mol

3) Determine the molar mass:

3.80 g / 0.031719 mol = 119.8 g/mol

4) The calculated answer is approximately double the known molar mass (60.0 g/mol) of acetic acid. What acetic acid does in benzene is to form dimers, composed of two molecules of acetic acid chemically joined together. This is the proposed structure for the dimer: 5) The formation of dimers would result in the van 't Hoff factor being amended to a value of 0.5.

Bonus Problem #2: What volume of ethylene glycol (C2H6O2), a nonelectrolyte, must be added to 20.0 L of water to produce an antifreeze solution with a freezing point of −34.0 °C? (The density of ethylene glycol is 1.11 g/cm3, and the density of water is 1.00 g/cm3.)

Solution:

1) We need to determine the moles of ethylene glycol:

Δt = i Kf m

34.0 °C = (1) (1.86 °C kg mol¯1) (x / 20.0 kg)

34.0 °C = (0.093 °C mol¯1) (x)

x = 365.59 mol

2) How many grams is this?

(365.59 mol) (62.0674 g/mol) = 22691.22 g

3) What volume does this occupy?

22691.22 g / 1.11 g/cm3 = 20442 cm3

This is 20.44 L

An approximate 50/50 (by volume) mixing of ethylene glycol and water gives us the freezing point depression we require. Note that I did not bother to show how 20.0 L of water was converted into 20.0 kg.

Bonus Problem #3: A compound contains 42.9% C, 2.4% H, 16.6% N, and 38.1% O by mass. The addition of 3.16 g of this compound to 75.0 mL of cyclohexane (d= 0.779 g/mL) gives a solution with a freezing point of 0.0 °C. The normal freezing point of cyclohexane is 6.5 °C and its freezing point depression constant is 20.2 °C/m. What is the molecular formula of the solute?

Solution:

1) Determine mass of cyclohexane:

75.0 mL x 0.779 g/mL = 58.425 g

2) Determine moles of compound dissolved:

Δt = i Kf m

6.5 °C = (1) (20.2 °C kg mol¯1) (x / 0.058425 kg)

6.5 °C = (345.7424 °C mol¯1) (x)

x = 0.0188 mol

3) Determine molecular weight of compound:

3.16 g / 0.0188 mol = 168 g/mol

4) Determine empirical formula of compound:

Assume 100 g of compound is present.

Convert masses to moles:

C ---> 42.9 g / 12.011 g/mol = 3.57 mol
H ---> 2.4 g / 1.008 g/mol = 2.38 mol
N ---> 16.6 g / 14.007 g/mol = 1.185 mol
O ---> 38.1 g / 16.00 g/mol = 2.38 mol

Divide through by the smallest:

C ---> 3.57 mol / 1.185 mol = 3
H ---> 2.38 mol / 1.185 mol = 2
N ---> 1.185 mol / 1.185 mol = 1
O ---> 2.38 mol / 1.185 mol = 2

The empirical formula is C3H2NO2

5) Determine the molecular formula:

C3H2NO2 weighs 84

168 / 82 = 2

The molecular formula is C6H4N2O4