What will be the freezing point of an aqueous solution containing 80.0 g NaCl in 3.4L of H2O?
Answers
Answer:
Explanation:
A
is due to getting the wrong sign on
Δ
T
f
.
C
is due to using
K
b
instead of
K
f
and getting the wrong sign.
D
is due to using
K
b
instead of
K
f
and getting the right sign.
INITIAL PREDICTIONS
First off, let's predict the freezing point from thinking about this qualitatively. Pure ethylene glycol has a freezing point of
−
12.9
∘
C
, and water's freezing point is
0
∘
C
.
So, the solution's freezing point should actually be below
0
∘
C
(what occurs is freezing point depression due to colligative properties of adding solutes into a solvent, so the freezing point should drop).
We can eliminate all but
B
,
−
20.1
∘
C
. That should be our answer before doing any work at all.
FREEZING POINT DEPRESSION FORMULA
Now let's actually calculate it so we can prove it. The High School version of the formula for freezing point depression is:
Δ
T
f
=
T
f
−
T
*
f
=
K
f
⋅
m
⋅
i
where:
T
f
is the freezing point of the solution.
T
*
f
is the freezing point of the pure solvent.
K
f
is the freezing point depression constant of the pure solvent.
m
is the molal concentration of the solution, which is the
mol
s of solute per
kg
of the pure solvent.
i
is the van't Hoff factor, which for ideal solutions is equal to the number of ions that dissociate in solution per formula unit.