Question

The experimentally measured freezing point of a 1.05 m aqueous solution of AlCl3 is -6.25°C. The freezing point depression constant for water is Kf = 1.86°C/m. Assume the freezing point of pure water is 0.00°C.

Part 1 What is the value of the van't Hoff factor for this solution?

Part 2 What is the predicted freezing point if there were no ion clustering in the solution?

Answer #1

**ANSWER:**

Given,

molality, m = 1.05 m

K_{f} = 1.86 °C/m

freezing point of pure water, T_{f}^{o} = 0.00
^{o}C

freezing point of aqueous solution of AlCl_{3},
T_{f} = -6.25°C

T_{f}
= T_{f}^{o} - T_{f} = 0.00 °C - (-6.25°C) =
6.25 °C

**Part 1:**

we know that,

T_{f}
= i x K_{f} x m

where,

i = van't Hoff factor

6.25°C = i x (1.86 °C/m) x 1.05 m

**i = 3.2**

**Hence, the value of the van't Hoff factor for this
solution is 3.2 .**

**Part 2:**

if there were no ion clustering in the solution, then vant's hoff factor is equal to 1.

T_{f}
= i x K_{f} x m

T_{f}
= K_{f} x m

T_{f}
= (1.86 °C/m) x 1.05 m = 1.95 ^{o}C

we know that,

T_{f}
= freezing point of pure water - freezing point of aquous
AlCl_{3} solution

T_{f}
= T_{f}^{o} - T_{f}

T_{f} =
T_{f}^{o} -
T_{f} = 0.00 ^{o}C - 1.95 ^{o}C = - 1.95
^{o}C

freezing point of aquous
AlCl_{3} solution = -1.95 ^{o}C

**Hence, freezing point of aquous solution of
AlCl _{3} is -1.95 ^{o}C, if there were no ion
clustering in the solution.**

1.The freezing point of an aqueous solution prepared by adding
0.0100 mol of acetic acid to 100. g of water is -0.190 C. The
freezing point depression of pure water is 0.000 C, and the
freezing point depression constant for water is 1.86 C/m. What is
the value for the van't Hoff factor for acetic acid in the aqueous
solution. You must show work to support your response.
2. Which of the following aqueous solutions should have the
lowest freezing...

The boiling point of an aqueous solution is 101.88 °C. What is
the freezing point? Constants can be found here.
Constants for freezing-point depression and boiling-point
elevation calculations at 1 atm:
Solvent
Formula
Kf value*
(°C/m)
Normal freezing
point (°C)
Kb value
(°C/m)
Normal boiling
point
(°C)
water
H2O
1.86
0.00
0.512
100.00
benzene
C6H6
5.12
5.49
2.53
80.1
cyclohexane
C6H12
20.8
6.59
2.92
80.7
ethanol
C2H6O
1.99
–117.3
1.22
78.4
carbon
tetrachloride
CCl4
29.8
–22.9
5.03
76.8
camphor
C10H16O...

What is the normal boiling point of an aqueous solution that has
a freezing point of 1.04 oC. Kf for water 1.86 oC/m (oC-kg/mol).
Hint: Calculate the molality from the freezing point depression and
use it to calculate the normal boiling point.

To use freezing-point depression or boiling-point elevation to
determine the molal concentration of a solution.
The freezing point, Tf, of a solution is lower than the
freezing point of the pure solvent. The difference in freezing
point is called the freezing-point depression, ΔTf:
ΔTf=Tf(solvent)−Tf(solution)
The boiling point, Tb, of a solution is higher than the
boiling point of the pure solvent. The difference in boiling point
is called the boiling-point elevation, ΔTb:
ΔTb=Tb(solution)−Tb(solvent)
The molal concentration of the solution, m,...

Calculate the freezing point of a solution containing 30 grams of
KCL and 3300.0 grams of water. The molal-freezing-point-depression
constant (Kf) for water is 1.86 degrees Celsius/M

Assuming complete dissociation of the solute, how many grams of
KNO3 must be added to 275 mL of water to produce a solution that
freezes at −14.5 ∘C? The freezing point for pure water is 0.0 ∘C
and Kf is equal to 1.86 ∘C/m ANSWER=109 g PART B NEEDED: If the
3.90 m solution from Part A boils at 103.45 ∘C, what is the actual
value of the van't Hoff factor, i? The boiling point of pure water
is 100.00...

what is the freezing point of an aqueous 200m (NH4)PO4 salt
solution? The Kf of water 1.86 C/m. Assume complete dissociation of
the soluble salt.

1) What is the freezing point of
an aqueous solution made by dissolving 0.139 mole of
MgCl2 in 100.0 g
of water? Kf =
1.86 oC/m

A solution of 5.00 g of sodium chloride in 1.00 kg of water has
a freezing point of –0.299°C. What is the actual (experimental) van
’t Hoff factor for this salt at this concentration?
Kf(water) =
1.86°C/m

Calculate the freezing point and boiling point of aqueous 1.9 m
CuCl3 given Kf for water = 1.86 deg.C/m; Kb for water = 0.512 deg
C/m. Assume theoretical value for i. Show work for credit.

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 19 minutes ago

asked 33 minutes ago

asked 48 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 2 hours ago

asked 2 hours ago

asked 2 hours ago