how many moles of sucrose should be dissolved in 500 grams of water so as to get a solution with a difference of 106 degree Celsius between boiling point and freezing point
Answers
Answer:
I got
72.2 g sucrose
.
What percent of the
5
∘
C
expansion of the transition temperature range is contributed by the boiling point elevation? (You should get about
22
%
.)
Well, boiling point elevation and freezing point depression are given by...
Δ
T
b
=
T
b
−
T
*
b
=
i
K
b
m
Δ
T
f
=
T
f
−
T
*
f
=
−
i
K
f
m
where
T
t
r
is the phase transition temperature of the solvent in the context of the solution, and
*
indicates pure solvent.
i
is the van't Hoff factor, i.e. the effective number of dissociated particles per solute particle placed into solvent.
K
b
=
0.512
∘
C
⋅
kg/mol
is the boiling point elevation constant.
K
f
=
1.86
∘
C
⋅
kg/mol
is the freezing point depression constant.
m
=
mols solute
kg solvent
is the molality of the solution.
Any solute added to a solvent will expand its boiling point to freezing point range.
If you want the gap to be
105
∘
C
between the new
T
f
and
T
b
, then...
T
b
−
T
f
=
105
∘
C
=
Δ
T
b
+
T
*
b
−
(
Δ
T
f
+
T
*
f
)
=
Δ
T
b
−
Δ
T
f
+
T
*
b
−
T
*
f
We know that the boiling point and freezing point of water are
100
∘
C
and
0
∘
C
respectively, so...
105
∘
C
=
Δ
T
b
−
Δ
T
f
+
100
∘
C
or
5
∘
C
=
Δ
T
b
−
Δ
T
f
=
i
K
b
m
+
i
K
f
m
=
i
m
(
K
b
+
K
f
)
Therefore, as sucrose is a nonelectrolyte,
i
=
1
and the required molality of the solution is:
m
=
5
∘
C
i
(
K
b
+
K
f
)
=
5
∘
C
1
⋅
(
0.512
∘
C
⋅
kg/mol
+
1.86
∘
C
⋅
kg/mol
)
=
2.108 mols solute/kg solvent
And since we have
100 g solvent
...
100 g water
→
0.100 kg water
Thus, the mols of sucrose in the water is:
2.108 mols solute
kg solvent
×
0.100
kg solvent
=
0.2108 mols sucrose
And this mass is
w
sucrose
=
0.2108
mols C
12
H
22
O
11
×
342.295 g sucrose
1 mol sucrose
=
72.2 g sucrose