1. Introduction to Molarity
Advanced Chemistry
Introduction to Molarity
Ms. Grobsky

2. Real-Life Applications of Concentration and Molarity
Prescription drugs in the correct concentration make you better
In higher concentration, they can kill you
Pesticides must be in proper concentrations
Food additives must be in correct concentrations
A driver is legally impaired at 0.08 mg/mL blood alcohol content

3. Chemistry and Molarity
For these reasons, chemists need to make solutions that have precise concentrations
But, how do they do that?
Chemists control the concentration of chemicals using the concept of molarity

4. Molarity and Solutions
Scientists use molarity to describe concentrations in chemistry
The concentration of a solution tells you how much solute that is dissolved in a given amount of solution (solute + solvent).
The molarity is the concentration of a solution
Molarity = Moles of solute
Liter of solution

5. Molarity
Conversion factors can be used to convert between amounts of solute and the solvent
Molarity does NOT mean number of moles!

6. ALWAYS START WITH WHAT YOU KNOW!
Sample Problem
4.0 L of solution is prepared from moles of NaCl. Calculate the molarity.
ALWAYS START WITH WHAT YOU KNOW!

7. Sample Problem
4.0 L of solution is prepared from moles of NaCl. Calculate the molarity.
Molarity = moles
4.0 L
= 0.56 moles/liter
= 0.56 M (Molar)
where M means moles per 1 liter

8. ALWAYS START WITH WHAT YOU KNOW!
Practice problems
250. mL of solution is prepared from 5.00 g KOH. Calculate the molarity.
ALWAYS START WITH WHAT YOU KNOW!

9. Practice problems
250. mL of solution is prepared from 5.00 g KOH. Calculate the molarity.
Molarity = 5.00 g x 1 mole
56.11 g
0.250 L
= M

10. Practice problems
How many moles are there in 205. mL of a M solution?
Remember, 205 mL of M solution does NOT mean there are .172 moles!
ALWAYS START WITH WHAT YOU KNOW!

11. Practice problems
How many moles are there in 205. mL of a M solution?
0.205 L x moles = moles L

12. ALWAYS START WITH WHAT YOU KNOW!
Practice problems
How many grams NaCl are there in mL of M solution?
ALWAYS START WITH WHAT YOU KNOW!

13. Practice problems
How many grams NaCl are there in mL of M solution?
0.250 L x moles x g = 7.31 g
1 L mole

14. ALWAYS START WITH WHAT YOU KNOW!
Practice problems
How many grams of NaCl must be used to prepare a mL of M solution?
ALWAYS START WITH WHAT YOU KNOW!

15. Practice problems
How many grams of NaCl must be used to prepare a mL of M solution?
L x moles x g = g
1 L mole

16. Preparation of Solutions
I could ask you to make me a 2.25 M solution of HCl using 1.45 liters of water, and you could do it. How?
Give directions for the preparation of 2.50 L of a 1.34 M NaCl solution

17. Practice problem
Describe how you would prepare mL of a M solution of CoCl2.
Find moles of CoCl2
Convert to grams
Mass amount of grams of CoCl2 on a balance
Dissolve in water
Transfer to a 100 mL volumetric flask and fill to the line

18. Making Molar Solutions
From Liquids(More accurately, from stock solutions)

19. Making Molar Solutions from Liquids
Not all compounds are in a solid form
Acids are purchased as liquids (“stock solutions”)
Yet, we still need a way to make molar solutions of these compounds
The procedure is similar to before
Use pipette to measure moles (via volume)
Use volumetric flask to measure volume
Now we use the equation M1V1 = M2V2
1 is starting (concentrated conditions)
2 is ending (dilute conditions)

20. The Dilution Formula
Example: If we have 1 L of 3 M HCl, what is M if we dilute acid to 6 L?
M1 = 3 mol/L, V1 = 1 L, V2 = 6 L
M1V1 = M2V2, M1V1/V2 = M2
M2 = (3 mol/L x 1 L) / (6 L) = 0.5 M
Why does the formula work?
Because we are equating mol to mol:
V1 = 1 L
M1 = 3 M
V2 = 6 L
M2 = 0.5 M
M1V1 = 3 mol
M2V2 = 3 mol

21. Practice problems
Problem 1 What volume of 0.5 M HCl can be prepared from 1 L of 12 M HCl? M1 = 12 mol/L, V1 = 1 L, M2 = 0.5 L M1V1 = M2V2 M1V1/M2 = V2 V2 = (12 mol/L x 1 L) / (0.5 L) = 24 L

22. Practice problems
Problem 2 1 L of a 3 M HCl solution is added to 0.5 L of a 2 M HCl solution. What is the final concentration of HCl? (Hint: first calculate total number of moles and total number of L) # mol = (3 mol/L)(1 L) + (2 mol/L)(0.5 L) = 3 mol + 1 mol = 4 mol # L = 1 L L = 1.5 L # mol/L = 4 mol / 1.5 L = 2.67 mol/L

23. Practice problems
How many mL of a 14 M stock solution must be used to make 250 mL of a 1.75 M solution?
You have 200 mL of 6.0 M HF. What concentration results if this is diluted to a total volume of 1 L?
100 mL of 6.0 M CuSO4 must be diluted to what final volume so that the resulting solution is 1.5 M?

24. Practice problems
What concentration results from mixing 400 mL of 2.0 M HCl with 600 mL of 3.0 M HCl?
What is the concentration of NaCl when 3 L of M NaCl are mixed with 2 L of 0.2 M NaCl?
What is the concentration of NaCl when 3 L of 0.5 M NaCl are mixed with 2 L of water?

25. Practice problems
Water is added to 4 L of 6 M antifreeze until it is 1.5 M. What is the total volume of the new solution?
There are 3 L of 0.2 M HF L of this is poured out, what is the concentration of the remaining HF?

26. Dilution problems (1-6, 6 two ways)1.
M1 = 14 M, V1 = ?, M2 = 1.75 M, V2 = 250 mL
V1 = M2V2 / M1 = (1.75 M)(0.250 L) / (14 M)
V1 = L = mL 2.
M1 = 6 M, V1 = 0.2 L, M2 = ?, V2 = 1 L
M2 = M1V1 / V2 = (6 M)(0.2 L) / (1 L)
M2 = 1.2 M 3.
M1 = 6 M, V1 = 100 mL, M2 = 1.5 M, V2 = ?
V2 = M1V1 / M2 = (6 M)(0.100 L) / (1.5 M)
V2 = 0.4 L or 400 mL

27. Dilution problems (4 - 6)
4. # mol = (2.0 mol/L)(0.4 L) + (3.0 mol/L)(0.6 L)
= 0.8 mol mol = 2.6 mol
# L = 0.4 L L
# mol/L = 2.6 mol / 1 L = 2.6 mol/L
5. # mol = (0.5 mol/L)(3 L) + (0.2 mol/L)(2 L)
= 1.5 mol mol = 1.9 mol
# mol/L = 1.9 mol / 5 L = 0.38 mol/L
6. # mol = (0.5 mol/L)(3 L) + (0 mol/L)(2 L)
= 1.5 mol + 0 mol = 1.5 mol
# mol/L = 1.5 mol / 5 L = 0.3 mol/L
Or, using M1V1 = M2V2,
M1 = 0.5 M, V1 = 3 L, M2 = ?, V2 = 5 L

28. Dilution problems (7, 8) 7. M1 = 6 M, V1 = 4 L, M2 = 1.5 M, V2 = ?
V2 = M1V1 / M2 = (6 M)(4 L) / (1.5 M)
V2 = 16 L
8. The concentration remains 0.2 M, both volume and moles are removed when the solution is poured out. Remember M is mol/L. Just like the density of a copper penny does not change if it is cut in half, the concentration of a solution does not change if it is cut in half.