Question

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

Answer #1

**we have elevation in BP formula as dT = i x Kb x
m**

**where i = vantoff factor = 2 for AgNO3 (since AgNo3
gives 2 ions per complete dissociation)**

**m = molality = 3.37 , Kb = boiling point
constant = 0.512C/m**

**we get dT = 2 x0.512 C/m x 3.37
m**

**dT = 3.45 = Tb ( solution) - Tb ( solvent
water)**

** 3.45 = Tb ( solution) -
100**

**Tb solution = 103.45 C**

**2) now we have formula for depression of freezing point
dT = i x Kf x m**

**where Kf = freezing point constant = 1.86 C/m
,**

**dT = 2 x 1.86 C/m x 3.37**

**dT = 12.54 C = T( freexing point of solvent) - T (
freezing point of solution)**

**12.54 = 0 - T( solution)**

**Freezing point of solution = - 12.54 C**

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

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

Assuming 100% dissociation, calculate the freezing point and
boiling point of 2.99 m AgNO3(aq). Constants may be found here.
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
37.8
176

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 (TfTf) and boiling point
(TbTb) of 1.92 m SnCl4(aq)1.92 m SnCl4(aq).
Colligative constants
can be found in the chempendix.

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

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Calculate the boiling point of a solution of NaCl that has a
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