Concentration and Dilution
Objectives:
1. Determine the molarity of the solution from the moles of a particular substance
2. Determine the volume of a specified stock solution needed to prepare a given
solution.
3. Determine the volume of water added to make the solution
Concentration
When we talk about the solution instead of number of moles we describe concentration.
Concentration shows how many moles of the substance are present per unit volume.
(The unit of volume is dm3).
The unit for concentration we use is molarity which means moles solute per liter of solvent or
moles per liter denoted with a capital M.
Example 1:
2M =
2 𝑚𝑜𝑙𝑒𝑠
𝑙
Answer this question:
What does it mean if: 3 M??
If we want to know the molarity of solution from the grams of solute we just convert to moles
using the molar mass of the substance and divided by the total volume of the solution.
Example 2:
Andrea had 10 g NaOH and she mixed it with the water till she got 1 liter solution of NaOH.
What would be the molarity of solution made from 10 g NaOH?
Step 1: Convert 10 g NaOH to moles
10.0 𝑔 𝑁𝑎𝑂𝐻 𝑥
1 𝑚𝑜𝑙𝑒 𝑁𝑎𝑂𝐻
40 𝑔 𝑁𝑎𝑂𝐻
= 0.25 mole
Step 2: Divided by the the total volume of solution
0.25 𝑚𝑜𝑙𝑒 𝑁𝑎𝑂𝐻
M=
= 0.25 𝑀
1𝑙
Answer the question and show your calculation:
1. What’s the molarity of a solution made from 20 g CuSO4 in 200 cm3 of water? (2)
Dilution
We can alter the concentration of a solution by a process called dilution. This is the act of
adding solvent (water) to a solution. We use the formula:
M1V1=M2V2
Which are:
M1: Molarity known
V1: Volume the known solution (will be taken from the known concentration, stock solution)
M2: Molarity the new solution
V2: Volume the new solution
Example 3:
To what volume would you need to dilute to make 1.95 M Ca(OH)2 from 5.85 M Ca(OH)2
with a 0.2 L solution.
Solution:
M1V1=M2V2
5.85 x 0,20 = 1.95 x V2
V2= 0.60 L
Answer the question and show your calculation:
1. If you dilute 200 cm3 of a 1.6 M solution of KCl to 1.0 dm3,
a. determine the new concentration of the solution. (2)
b. explain how do you make the new solution/new concentration in the lab. (2)
2. How many milliliters of 2.0 M barium sulfate solution must be added to 160 mL of
water to achieve a 0.20 M barium sulfate solution? (2)