### BP elevation and FP depression asked in the same question(with some osmosis too)

Problem #1: The boiling point of an aqueous solution is 101.21 °C. What is the freezing point?

Solution:

1) Determine the molality of the solution:

Δt = i Kb m

1.21 °C = (1) (0.52 °C kg mol-1) (x)

x = 2.326923 m

2) Determine the freezing point depression with the above molality:

Δt = i Kf m

y = (1) (1.86 °C kg mol-1) (2.326923 mol/kg)

y = 4.33 °C

The freezing point of this solution is -4.33 °C.

Note that I assumed the solute was a non-electrolyte, producing a van 't Hoff factor of 1. I could have used any factor value I wanted. The bp factor winds up in the denominator and the fp factor is used in the numerator. They cancel out.

Problem #2: An aqueous solution of an unknown solute freezes at -3.55 °. At what temperature would you expect it to boil?

Solution:

1) We must determine the molality of the solution:

Δt = i Kf m

3.55 °C = (1) (1.86 °C kg mol-1) (x)

x = 1.9086 m

2) Determine the boiling point elevation with the above molality:

Δt = i Kb m

y = (1) (0.52 °C kg mol-1) (1.9086 mol/kg)

y = 0.99 °C

The boiling point of this solution is 100.99 °C.

Problem #3: How many grams of sucrose should be added to 500.0 grams of water so that the difference in freezing point and boiling point is 105.0 °C

Solution:

Δt1 = i Kf m
Δt2 = i Kb m

The sum of Δt1 and Δt2 is 5.0 °C.
The van 't Hoff factor of sucrose is 1.
The molality is x / 0.5

Δt1 + Δt2 = (Kf m) + (Kb m)

5 = (Kf m) + (Kb m)

5 = (1.86 * (x / 0.5)) + (0.52 * (x / 0.5))

5 = 3.72x + 1.04x

4.76x = 5

x = 1.05 mol

1.05 mol times 342.2948 g/mol = 360. g

Problem #4: Calculate the freezing point and the boiling point of a solution that contains 15.0 grams of urea (CH4N2O) in 250. grams of water. Urea is a covalently bounded compound.

Solution:

1) Moles of urea:

15.0 g / 60.06 g/mol = 0.24975 mol

2) Freezing point calculation:

Δt = i Kf m

x = (1) (1.86 °C kg mol-1) (0.24975 mol / 0.250 kg)

x = 1.86 °C

The freezing point is -1.86 °C

3) Boiling point calculation:

Δt = i Kb m

x = (1) (0.52 °C kg mol-1) (0.24975 mol / 0.250 kg)

x = 0.52 °C

The boiling point is 100.52 °C

Problem #5: A solution is 30.5% by mass of antifreeze (C2H6O2) in water. For this solution:

a) calculate the freezing point depression and the freezing point
b) calculate the boiling point elevation and the boiling point.

Solution:

1) Moles of antifreeze:

30.5% means 30.5 g of C2H6O2 in 100 g of solution

30.5 g / 62.0674 = 0.4914 mol

2) FP depression:

x = (1) (1.86 °C kg mol-1) (0.4914 mol / 0.0695 kg)

x = 13 °C (leading to a freezing point of -13 °C)

The mass of water came from 100 g - 30.5 g = 69.5 g = 0.0695 kg

3) BP elevation:

x = (1) (0.52 °C kg mol-1) (0.4914 mol / 0.0695 kg)

x = 3.68 °C (leading to a boiling point of 103.68 °C)

Note: the substance is a molecular one, leading to a van 't Hoff factor of 1.

Problem #6: What is the approximate osmotic pressure of a 0.118 m solution of LiCl at 10.0 °C? The freezing point of this solution is -0.415 °C.

Solution:

1) Use the freezing point data to determine the van 't Hoff factor:

0.415 = (i) (1.86) (0.118)

i = 1.89

2) Determine the approximate osmotic pressure:

π = (1.89) (0.118) (0.08206) (283)

π = 5.18 atm

Note: the answer is approximate because we assume a density of 1.00 g/mL for the solution. This allowed the 0.118 molal solution to be treated as 0.118 molar. A density of 1.00 g/mL is a reasonable one, but the assumption does result in the answer being labeled approximate.

Problem #7: At 286 K, the osmotic pressure of a glucose solution is 9.97 atm. What is the freezing point depression (given the density of the solution is 1.12 g/mL)?

Solution:

1) Determine the molarity of the solution:

9.97 atm = (1) (M) (0.08206 L atm/mol K) (286 K)

M = 0.4248128 mol/L

2) Convert molarity to molality:

Assume 1.00 L of the solution is present.

1000 mL times 1.12 g/mL = 1120 g <--- mass of the 1.00 L of solution

The solution contains 0.4248128 mol of glucose

0.4248128 mol times 180.16 g/mol = 76.5343 g

1120 g - 76.5343 g = 1043.466 g <--- mass of the water in the 1.00 L of solution

0.4248128 mol / 1.043466 kg = 0.407117 molal

3) Determine the freezing point depression:

Δt = (1) (1.86) (0.407117)

Δt = 0.76 °C

Problem #8: What are the normal freezing points and boiling points of 21.2 g of CaCl2 in water?

Solution:

The key lies in the word normal. The normal boiling point of water is 100 °C and the normal freezing point of water is 0 °C. A solution has a boiling point and a melting point different from the normal values, but the normal values themselves remain at 0 and 100.

Is this a trick question? Remember, any question that you do not know the answer to is a trick question.

Problem #9: An aqueous solution of KI has a freezing point of -1.95 °C and an osmotic pressure of 25.0 atm at 25.0 °C. Assuming that the KI completely dissociates in water, what is the density of the solution?

Solution:

1) Solve for the molality:

Δt = iKfm

1.95 °C = (2) (1.86 °C/m) (x)

x = 0.52419 m

2) Solve for the molarity:

π = iMRT

25.0 atm = (2) (y) (0.08206 L atm / mol K) (298 K)

y = 0.511166 M

3) 1.000 L of the solution contains 0.511166 moles of KI and is 0.52419 m. Solve for the mass of water:

0.52419 molal = 0.511166 moles / z

z = 0.975154 kg = 975.154 g

4) Solve for the mass of KI:

0.511166 mol times 165.998 g/mol = 84.8525 g

5) Compute the density:

975.154 g + 84.8525 g = 1060.0065 g (this is the mass of 1.000 L of solution)

1060.0065 g / 1000. mL = 1.060 g/mL