1. A measurement of the concentration of a solution
MOLARITY
A measurement of the concentration of a solution
Molarity (M) is equal to the moles of solute (n) per liter of solution M = n / V = mol / L
Calculate the molarity of a solution prepared by mixing 1.5 g of NaCl in mL of water.
First calculate the moles of solute:
1.5 g NaCl (1 mole NaCl) = moles of NaCl
58.45 g NaCl
Next convert mL to L: L of solution
Last, plug the appropriate values into the correct variables in the equation:
M = n / V = moles / L = mol/L

2. MOLARITY M = n / V , now rearrange to solve for n: n = MV
M = n / V = 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 = n / V , now rearrange to solve for n: n = MV
n = (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 (23.95 g LiOH / 1 mol LiOH) = 7.48 g of LiOH

3. 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 therefore:
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
Next convert mL to L:
mol/mL (1000 mL/ 1L) = 5.51 mol/L = 5.51 M

4. 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 (nb) and the moles of solute after the dilution (na) are the same: nb = na
And the moles for any solution can be calculated by n=MV
A relationship can be established such that
MbVb = nb = na = MaVa
Or simply : MbVb = MaVa

5. 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

6. 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 about 100 mL of DI water in a mL graduated cylinder then more water would be added to the mixture until the bottom of the menicus is at mL. Mix well.

7. PRACTICE PROBLEMS
2.4 M
_________1. What is the concentration of mL of 0.60 moles of HCl?
_________ 2. What is the concentration of 35.0 mL of moles of KCl?
_________ 3. How many grams of KCl is needed to prepare 50.0 mL of a 0.10 M solution?
_________ 4. How many milliliters of water must be added to 30.0 mL of 9.0 M KCl to make a solution that is 0.50 M KCl?
_________ 5. What volume of M LiOH will contain 55.3 g of LiOH?
_________ 6. How many liters of water must be added to mL of 4.50 M HBr to make a solution that is M HCl?
1.59 M
0.38 g
510 mL
3.00 L
1.80 L