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1 mol CaF 2

1.5 atom F X atoms F 1,000,000 m’cule H 2 O 3.34 x 10 28 m’cule H 2 O. =. What mass of CaF 2 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 x10 23 m’cule.

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1 mol CaF 2

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  1. 1.5 atom F X atoms F 1,000,000 m’cule H2O 3.34 x 1028 m’cule H2O = 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 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

  2. mol M L How many moles solute are required to make 1.35 L of 2.50 M solution? = mol = M L 3.38 mol = 2.50 M (1.35 L) 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

  3. mol M = L 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? 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. mass of solvent only 1 kg water = 1 L water Molality

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