Solubility of potassium nitrate solution at room temperature (30°C) is 60 g/ 100 g of water. 240 g of saturated solution of potassium nitrate are allowed to evaporate such that 80 g of water evaporates. Find the amount of crystals deposited due to evaporation.
Answers
Answer:
Explanation:
So, you know that potassium nitrate,
KNO
3
, has a solubility of
155 g
per
100 g
of water at
75
∘
C
and of
38.0 g
at
25
∘
C
.
What that means is that at
75
∘
C
, you can dissolve as much as
155 g
of solid potassium nitrate in water without the solution becoming saturated.
Once you hit that
155 g
mark, the solution becomes saturated, which means that the solution can't dissolve any more solid.
Now, the same thing can be said for the solution at
25
∘
C
. At this temperature, dissolving less than
38.0 g
of potassium nitrate per
100 g
of water will result in an unsaturated solution.
At the
38.0 g
mark, the solution will become saturated.
So, you are starting with
100 g
of saturated solution at
75
∘
C
. Mind you, you have
100 g
of solution that contains as much dissolved potassium nitrate as possible.
This solution will contain
100.0
g solution
⋅
155 g KNO
3
(
155
+
100
)
g solution
=
60.78 g KNO
3
Now, potassium nitrate's solubility is given per
100 g
of water. Calculate how much water you have in this initial solution
m
water
=
100.0 g
−
60.78 g
=
39.22 g
Next, determine how much potassium nitrate can be dissolved in
39.22 g
of water at
26
∘
C
in order to make a saturated solution, i.e. have the maximum amount of dissolved potassium nitrate possible
39.22
g water
⋅
38.0 g KNO
3
100
g water
=
14.9 g KNO
3
This means that when the initial solution is cooled from
75
∘
C
to
25
∘
C
, the amount of water that it contained will only hold
14.9 g
of dissolved potassium nitrate.
The rest will crystallize out of solution
m
crystallize
=
60.78 g
−
14.9 g
=
45.9 g
The answer is rounded to three sig figs.
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