Answer:
The problem provides you with the solubility of potassium chloride,
KCl
, in water at
20
∘
C
, which is said to be equal to
34 g / 100 g H
2
O
.
This means that at
20
∘
C
, a saturated solution of potassium chloride will contain
34 g
of dissolved salt for every
100 g
of water.
As you know, a saturated solution is a solution that holds the maximum amount of dissolved salt. Adding more solid to a saturated solution will cause the solid to remain undissolved.
In your case, you can create a saturated solution of potassium chloride by dissolving
34 g
of salt in
100 g
of water at
20
∘
C
.
Now, your goal here is to figure out how much potassium chloride can be dissolved in
300 g
of water at this temperature. To do that, use the given solubility as a conversion factor to take you from grams of salt to grams of water
300
g H
2
O
⋅
34 g KCl
100
g H
2
O
=
102 g KCl
You should round this off to one sig fig, since that is how many sig figs you have for the mass of water
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
mass of KCl
=
100 g
a
a
∣
∣
−−−−−−−−−−−−−−−−−−−−−−−
Explanation:
