1. Molarity and all that is Molarity (well not all, but a lot)
Solution Chemistry
Molarity and all that is Molarity
(well not all, but a lot)

2. SOLUTION CHEMISTRY Concentration:
amount of solute in a given amount of solvent (can be determined quantitatively)
Dilute:
a solution with a small amount of solute per solvent amount (relative term)
Concentrated:
a solution with a large amount of solute per solvent amount (relative term)
BOTH DILUTE AND CONCENTRATED ARE QUALITATIVE

3. Concentration of Solute
The amount of solute in a solution is given by its concentration.
Molarity ( M ) =
moles solute
liters of solution
The concentration of a solution is said to be its molarity.
Ex. 1 M CuSO4
“1 molar copper II sulfate

4. MOLARITY M = mol solute L sol’n M = 4.67 mol Li2SO3 2.04 L Li2SO3
A measurement of the concentration of a solution
Molarity (M) is equal to the moles of solute per liter of solution
M = mol / L
Determine the Molarity (M) of the following solution. You have 4.67 moles of Li2SO3 and you want to make 2.04 L of the solution.
M = mol solute
L sol’n
M = mol Li2SO3
2.04 L Li2SO3

5. MOLARITY M = mol of solute / L sol’n Calculate the moles of solute:
A measurement of the concentration of a solution
Molarity (M) is equal to the moles of solute per liter of solution
M = mol of solute / L sol’n
Calculate the molarity of a solution prepared by mixing 1.5 g of NaCl in .50 L of water.
Calculate the moles of solute:
1.5 g NaCl X 1 mol NaCl =
58.45 g NaCl
Plug the appropriate values into the correct variables in the equation:
M = mole solute/L sol’n = moles / L = mol/L
mol NaCl

6. MOLARITY M = mol / L
How many grams of LiOH is needed to prepare mL of a 1.25 M solution?
First calculate the moles of solute needed:
M = mole solute  rearrange to solve for mole solute:
L sol’n
Change mL to L of solution
Plug in the variables in the formula
mole solute = (1.25 mol / L) ( L) = moles of solute needed
Next calculate the molar mass of LiOH: g/mol
Last, use diminsional analysis to solve for mass:
moles x g LiOH / 1 mol LiOH) = 7.48 g of LiOH

7. MOLARITY M = n / V = mol / L
What is the molarity of hydroiodic acid if the solution is 47.0% HI by mass and has a density of 1.50 g/mL?
First calculate the mass of solute in the 47.0% solution using the density. The 1.50 g/mL is the density of the solution but only 47.0% of the solution is the solute
47.0% of 1.50 g/mL = (0.470) (1.50 g/mL) = g/mL density of solute
Since molarity is given in moles per liter and not grams we must convert the g/mL to mol/mL using the molar mass.
0.705 g/mL (1 mole/ 128 g) = mol/mL
Convert mL to L:
mol/mL (1000 mL/ 1L) = 5.51 mol/L = 5.51 M

8. MOLARITY & DILUTION M1V1 = M2V2
The act of diluting a solution is to simply add more water (the solvent) thus leaving the amount of solute unchanged.
Since the amount or moles of solute before dilution (n1) and the moles of solute after the dilution (n2) are the same: n1 = n2
And the moles for any solution can be calculated by n=MV
A relationship can be established such that
M1V1 = n1 = n2 = M2V2
Or simply : M1V1 = M2V2

9. M1 V1 = M2 = (0.05 mol/L) (25.0 mL) = 0.0167 M of KI
MOLARITY & DILUTION
Calculate the molarity of a solution prepared by diluting 25.0 mL of 0.05 M potassium iodide with 50.0 mL of water (the densities are similar).
M1 = 0.05 mol/L M2 = ?
V1 = 25.0 mL V2 = = 75.0 mL
M1V1 = M2V2
M1 V1 = M2 = (0.05 mol/L) (25.0 mL) = M of KI
V mL

10. M2 V2 = V1 = (0.150 mol/L) (250.0 mL) = 6.25 mL of 6 M HCl
MOLARITY & DILUTION
Given a 6.00 M HCl solution, how would you prepare mL of M HCl?
M1 = 6.00 mol/L M2 = 0.150
V1 = ? mL V2 = mL
M1V1 = M2V2
M2 V2 = V1 = (0.150 mol/L) (250.0 mL) = mL of 6 M HCl
M mol/L
You would need 6.25 mL of the 6.00 M HCl reagent which would be added to a mL graduated cylinder then more water would be added to the mixture until the bottom of the menicus is at mL. Mix well.