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4.3 Solutions HW
CLASSIFYING SOLUTIONS - Classify the following solutions as
Unsaturated, Saturated or Supersaturated. For e) and f), you will need to
use a ratio to calculate the number of grams of solute in 100 grams of
water.)
38 g of NaCl in 100 grams of water at 100˚ C Below line Unsaturated
120 g KI in 100 grams of water at 30˚ C Below line Unsaturated
120 g KNO3 in 100 grams of water at 60˚ C On line Saturated
50 g NaCl in 100 grams of water at 60˚ C Above line Super Saturated
100 g NaCl in 200 grams of water at 60˚ C
The Equivalent Ratio is 50 g in 100 grams of water (Same as above)
140 g NaNO3 in 200 grams of water at 15˚ C ______
The Equivalent Ratio is 70 g in 100 grams of water - Below line
Unsaturated
Which Compound’s solubility is least changed by a change in
temperature? NaCl has a graph that is nearly constant.
Which Compound’s solubility is most changed by a change in
temperature? KNO3 has the steepest graph.
Which compound is the most soluble? KI, the highest line.
Which compound is less soluble at higher temperatures (this does not
follow the general rule) ? NH3 and Ce2(SO4)3
What is the significance of the graphs for two substances crossing
(for example, NH3 and KCLO3 cross at approximately 56 ˚ C) ?
At this temperature the two substances have the same solubility.
𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 =
# 𝑜𝑓 𝑀𝑜𝑙𝑒𝑠 𝑜𝑓 𝑆𝑜𝑙𝑢𝑡𝑒
𝑙𝑖𝑡𝑒𝑟𝑠 𝑜𝑓 𝑆𝑜𝑙𝑣𝑒𝑛𝑡
Dilution/Evaporation Questions –
1. 2 liters of a .5 molar solution are diluted by adding 2 liters of
water. What is the final concentration of the diluted solution?
First find the number of moles in the original solution:
moles = Molarity x liters = (. 5
𝑚𝑜𝑙𝑒𝑠
) ( 2 𝑙𝑖𝑡𝑒𝑟) = 1 mole
𝑙𝑖𝑡𝑒𝑟
Now calculate the concentration of the diluted solution:
m = 1 mole, 𝑙 = 2 𝑙𝑖𝑡𝑒𝑟𝑠 + 2 𝑙𝑖𝑡𝑒𝑟𝑠 = 4 𝑙𝑖𝑡𝑒𝑟𝑠
𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 =
# 𝑜𝑓 𝑀𝑜𝑙𝑒𝑠 𝑜𝑓 𝑆𝑜𝑙𝑢𝑡𝑒
1 𝑚𝑜𝑙𝑒
𝑚𝑜𝑙𝑒
=
= .25
𝑙𝑖𝑡𝑒𝑟𝑠 𝑜𝑓 𝑆𝑜𝑙𝑣𝑒𝑛𝑡
4 𝑙𝑖𝑡𝑒𝑟𝑠
𝑙𝑖𝑡𝑒𝑟
3. If 10 liters of a 3 molar solution are left out in the sun and 5 liters
of water evaporates, what will be the final concentration of the
solutions?
First find the number of moles in the original solution:
moles = Molarity x liters = (3
𝑚𝑜𝑙𝑒𝑠
𝑙𝑖𝑡𝑒𝑟
) ( 10 𝑙𝑖𝑡𝑒𝑟𝑠) = 30 moles
Now calculate the concentration of the diluted solution:
m = 30 moles, 𝑙 = 5 𝑙𝑖𝑡𝑒𝑟𝑠
𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 =
# 𝑜𝑓 𝑀𝑜𝑙𝑒𝑠 𝑜𝑓 𝑆𝑜𝑙𝑢𝑡𝑒
30 𝑚𝑜𝑙𝑒
𝑚𝑜𝑙𝑒
=
=6
𝑙𝑖𝑡𝑒𝑟𝑠 𝑜𝑓 𝑆𝑜𝑙𝑣𝑒𝑛𝑡
5 𝑙𝑖𝑡𝑒𝑟𝑠
𝑙𝑖𝑡𝑒𝑟
4. 146 grams of NaCl are dissolved in 2 liters of water. If the
desired concentration is 2.5 Molar, how much water must be added
or evaporated to attain the correct concentration?
First find the Molar Mass of NaCl (a skill you should have)
2. 2 liters of a 1.5 molar solution are diluted by adding 1 liter of
water. What is the final concentration of the diluted solution?
Molar Mass of NaCl = 58 grams/mole
Convert grams to moles (a skill you should have)
First find the number of moles in the original solution:
moles = Molarity x liters = (1.5
𝑚𝑜𝑙𝑒𝑠
𝑙𝑖𝑡𝑒𝑟
) ( 2 𝑙𝑖𝑡𝑒𝑟) = 3 moles
Now calculate the concentration of the diluted solution:
m = 1 mole, 𝑙 = 2 𝑙𝑖𝑡𝑒𝑟𝑠 + 1 𝑙𝑖𝑡𝑒𝑟𝑠 = 3 𝑙𝑖𝑡𝑒𝑟𝑠
𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 =
# 𝑜𝑓 𝑀𝑜𝑙𝑒𝑠 𝑜𝑓 𝑆𝑜𝑙𝑢𝑡𝑒
3 𝑚𝑜𝑙𝑒
𝑚𝑜𝑙𝑒
=
=1
𝑙𝑖𝑡𝑒𝑟𝑠 𝑜𝑓 𝑆𝑜𝑙𝑣𝑒𝑛𝑡
3 𝑙𝑖𝑡𝑒𝑟𝑠
𝑙𝑖𝑡𝑒𝑟
146 𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝑁𝑎𝐶𝑙
1 𝑚𝑜𝑙𝑒
×
= 2.5 𝑚𝑜𝑙𝑒𝑠
1
58 𝑔𝑟𝑎𝑚𝑠
Calculate the Molarity
𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 =
# 𝑜𝑓 𝑀𝑜𝑙𝑒𝑠 𝑜𝑓 𝑆𝑜𝑙𝑢𝑡𝑒
2.5 𝑚𝑜𝑙𝑒𝑠
𝑚𝑜𝑙𝑒
=
= 1.25
𝑙𝑖𝑡𝑒𝑟𝑠 𝑜𝑓 𝑆𝑜𝑙𝑣𝑒𝑛𝑡
2 𝑙𝑖𝑡𝑒𝑟𝑠
𝑙𝑖𝑡𝑒𝑟
Adjust Concentration
The molarity (concentration) is half of the desired concentration so
reducing the water by 50% (cutting it in half by evaporation) will
increase the concentration by a factor or 2 (or twice )
5. A person wants to make a concentrated drink mix (like lemonade)
to be diluted later. The correct concentration of sugar in the
lemonade according to the recipe is 33 grams of sugar in 100 grams
of water. When diluted the goal is to have 20 liters (20,000 grams)
of lemonade. The goal is to make a 2X (double) concentration for
later dilution. How much (or how many grams) of sugar are
required?
What is the Ratio According to the Recipe
Here it is easier to work in grams of solute per 100 grams of water.
This is the common unit of concentration for the solubility graphs.
According the recipe the ratio is:
33 𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝑆𝑢𝑔𝑎𝑟
100 𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝑊𝑎𝑡𝑒𝑟
Goal is to end up with 20 liters, How many grams is this?
20,000 grams
Create an Equivalent Ratio
33 𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝑆𝑢𝑔𝑎𝑟
𝑋 𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝑆𝑢𝑔𝑎𝑟
=
100 𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝑊𝑎𝑡𝑒𝑟 20,000 𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝑊𝑎𝑡𝑒𝑟
Solving gives 6,600 grams of sugar.
This is about 14.5 pounds
It turns out you did not need to know the dilution.