Question

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.

Answer #1

Given:

Freezing point of the solution = 1.04 ^{0}C

Kf (water) = 1.86 ^{0}C kg /mol

Step 1 : Calculation of molality

Delta Tf = m kf

m = delta Tf / kf

we know

Delta Tf = Freezing point of the solution – freezing point of the solvent.

Delta Tf = 1.04 deg C – 0 deg C

Delta Tf = 1.04 deg C

m = 1.04 deg C / 1.86 deg C kg /mol

= 0.55914 mol/Kg

Now we have to find the normal boiling point f the solution.

We know

Delta Tb = m kb

Here delta Tb is the elevation in boiling point , m is molality , kb is boiling point constant of solvent.

Kb for water = 0.512 deg C kg /mol

Lets plug the value of m and kb to calculate delta Tb

Delta Tb = 0.55914 x 0.512 = 0.28628 deg C

Delta Tb = boiling point of the solution – boiling point of pure solvent ‘

0.28628 deg C = boiling point of the solution – 100 deg C

Boiling point of the solution = 100 + 0.28628 = 100.3 deg C

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...

5. Determine the molal freezing point depression constant (Kf)
(in °C⋅kg/mol) of water by using the data of three NaCl solutions .
Use Equation 3 in the “Background and Procedure” file, the molality
values of all solutes from Question s 1 – 3 and the freezing point
depression ( ∆T) from Question 4. In your calculation, rewrite the
symbol“m ” (the molality) as “mol/kg” so that the molal freezing
point depression constant will have the correct unit. Make
sure to...

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

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...

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 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.

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?

Calculate the boiling point of a solution of NaCl that has a
freezing point of -0.3720 °C. Assume complete dissociation. Kf
water = 1.86 °C/m Kb water = 0.512 °C/m
A. 100.1 °C
B. 99.1 °C
C. 101.1 °C
D. 98.9 °C
E. 105 °C

What is the molality of NaCl formula units in an aqueous
solution if the boiling point of the solution is 102.10 °C at 1
atm. (Water has a boiling point elevation constant of 0.51 °C
kg/mol.)
A) 4.1 m
B) 0.49 m
C) 8.2 m
D) 2.1 m

Calculate the freezing point and boiling point of each of the
following solutions: the freezing point of the solution: 174 g of
sucrose, C12H22O11, a
nonelectrolyte, dissolved in 1.35 kg of water
(Kf=1.86∘C) Express your answer using one decimal
place.

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 35 minutes ago

asked 40 minutes 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

asked 2 hours ago

asked 2 hours ago

asked 3 hours ago

asked 3 hours ago