300grams of a substance 'x', has solubility of 35and 50 at 25°Cand50°C respectively.Temperature of solution is decreased from 50°C to 25°C.Find amount of 'x' that was seperated as a precipitate.
Answers
Answer:
45.9 g
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.
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