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 3.34 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 = 2.505 x 1022 molecules = 3.25 g CaF2 6.02 x 1023 molecules 1 mol CaF2

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 = 3.38 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 = 3.38 mol Mg3(PO4)2 = 889 g Mg3(PO4)2 1 mol Mg3(PO4)2

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 = 58.6 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 0.342 mol solute? M = mol L 0.342 mol M = = 0.061 M Ba(OH)2 5.65 L

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