1. What mass of CaF2 must be added to 1,000 L of water
so that fluoride atoms are present at a conc. of 1.5 ppm?
1000 mL
1 g
1 mol
6.02 x1023 m’cule
X m’cule H2O = 1000 L
= 3.34 x 1028 m’cules H2O
1 L
1 mL
18 g
1 mol
1.5 atom F X atoms F
1,000,000 m’cule H2O x 1028 m’cule H2O =
1 molecule CaF2
X = 5.01 x 1022 atoms F
times =
2.505 x 1022 molecules CaF2
2 atoms F
1 mol CaF2
78.1 g CaF2
X g CaF2 = x 1022 molecules =
3.25 g CaF2
6.02 x 1023 molecules
1 mol CaF2

2. mol L M
How many moles solute are required to make 1.35 L of 2.50 M solution?
mol = M L
= 2.50 M (1.35 L) =
3.38 mol
A. What mass sodium hydroxide is this?
40.0 g NaOH
X g NaOH = mol NaOH =
135 g NaOH
1 mol NaOH
B. What mass magnesium phosphate is this?
262.9 g Mg3(PO4)2
X g Mg3(PO4)2 = mol Mg3(PO4)2 =
889 g Mg3(PO4)2
1 mol Mg3(PO4)2

3. Find molarity if 58.6 g barium hydroxide are in 5.65 L solution.
Step 1). How many moles barium hydroxide is this?
1 mol Ba(OH)2
X mol Ba(OH)2 = g Ba(OH)2 =
0.342 mol Ba(OH)2
171.3 g Ba(OH)2
Step 2). What is the molarity of a 5.65 L solution containing mol solute?
M =
mol L
0.342 mol
M = =
0.061 M Ba(OH)2
5.65 L

4. You have 10.8 g potassium nitrate.
How many mL of solution will make this a 0.14 M solution?
convert to mL

5. Molality
mass of solvent only
1 kg water = 1 L water