Question

Assuming 100% dissociation, calculate the freezing point and boiling point of 1.99 m AgNO3(aq). Constants may be found here.

Answer #1

[AgNO3]= 1.99 m , we know m is molality which is equal to = mol of solute / Mass of solute in kg.

Since the dissociation is 100 % so the i ( number of particles that are formed in the solution ) = 2

Solution:

We use following formulae.

Delta Tf = i x m x kf

Here I = 2 , m is molality, kf is the freezing point constant of water.

Delta Tf = freezing pint of the solvent – freezing point of the
solution = 0^{0} C – (freezing point of the solution)

Kf of water = -1.86 ^{0}C/ m

Lets plug the given values.

-(freezing point of the solution) = 2 x 1.99 m x ( -1.86
^{0}C/m)

**Freezing point of the solution =- 7.40
^{0}C**

Assuming 100% dissociation, calculate the freezing point and
boiling point of 2.19 m Na2SO4(aq). Constants may be found
here.

Assuming 100% dissociation, calculate the freezing point and
boiling point of 1.52 m SnCl4(aq). Constants may be found here.
vent
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
37.8
176

Assuming 100% dissociation, calculate the freezing point and
boiling point of 3.37 m AgNO3(aq).

Assuming 100% dissociation, calculate the freezing point and
boiling point of 1.19 m K3PO4(aq)

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...

Calculate the freezing point and boiling point in each solution,
assuming complete dissociation of the solute. Part A Calculate the
freezing point of a solution containing 12.3 g FeCl3 in 180 g
water. Tf = ∘C Request Answer Part B Calculate the boiling point of
a solution above. Tb = ∘C Request Answer Part C Calculate the
freezing point of a solution containing 4.2 % KCl by mass (in
water). Express your answer using two significant figures. Tf = ∘C...

The boiling point of an aqueous solution is 101.77 °C. What is
the freezing point? Constants can be found here.

Calculate the freezing point of the following solutions,
assuming complete dissociation. Part A 10.4 g FeCl3 in 151 g water
Part B 3.2 % KCl by mass (in water) Express your answer using two
significant figures. Part C 0.175 m MgF2 Part D Calculate the
boiling point of the solution in part A, assuming complete
dissociation. Part E Calculate the boiling point of the solution in
part B, assuming complete dissociation. Part F Calculate the
boiling point of 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

Calculate the freezing point and boiling point of a solution
containing 13.6 g of naphthalene (C10H8) in 110.0 mL of benzene.
Benzene has a density of 0.877 g/cm3. A)Calculate the freezing
point of a solution. (Kf(benzene)=5.12∘C/m.) B)Calculate
the boiling point of a solution.
(Kb(benzene)=2.53∘C/m.)

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