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Avogadro’s Number

Avogadro’s Number. Avogadro’s number is written as conversion factors. 6.02 x 10 23 particles and 1 mole 1 mole 6.02 x 10 23 particles The number of molecules in 0.50 mole of CO 2 molecules is calculated as 0.50 mole CO 2 molecules x 6.02 x 10 23 CO 2 molecules

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Avogadro’s Number

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  1. Avogadro’s Number • Avogadro’s number is written as conversion factors. 6.02 x 1023 particles and 1 mole 1 mole 6.02 x 1023 particles • The number of molecules in 0.50 mole of CO2 molecules is calculated as 0.50 mole CO2 molecules x 6.02 x 1023 CO2 molecules 1 mole CO2 molecules = 3.0 x 1023 CO2 molecules

  2. Learning Check A. Calculate the number of atoms in 2.0 moles of Al. 1) 2.0 Al atoms 2) 3.0 x 1023 Alatoms 3) 1.2 x 1024 Al atoms B. Calculate the number of moles of S in 1.8 x 1024 S. 1) 1.0 mole S atoms 2) 3.0 mole S atoms 3) 1.1 x 1048 mole S atoms

  3. Solution A. Calculate the number of atoms in 2.0 moles of Al. 3) 1.2 x 1024 Al atoms 2.0 moles Al x 6.02 x 1023 Al atoms 1 mole Al B. Calculate the number of moles of S in 1.8 x 1024 S. 2) 3.0 mole S atoms 1.8 x 1024 S atoms x 1 mole S 6.02 x 1023 S atoms

  4. Molar Mass of CaCl2 • For a compound, the molar mass is the sum of the molar masses of the elements in the formula. We calculate the molar mass of CaCl2 to the nearest 0.1 g as follows.

  5. Molar Mass of K3PO4 Determine the molar mass of K3PO4 to 0.1 g.

  6. Molar Mass Factors Methane CH4 known as natural gas is used in gas cook tops and gas heaters. 1 mole CH4 = 16.0 g The molar mass of methane can be written as conversion factors. 16.0 g CH4 and 1 mole CH4 1 mole CH4 16.0 g CH4

  7. Learning Check A glucose solution with a volume of 2.0 L contains 72 g glucose (C6H12O6). If glucose has a molar mass of 180. g/mole, what is the molarity of the glucose solution? 1) 0.20 M 2) 5.0 M 3) 36 M

  8. Solution 1) 0.20 M 72 g x 1 mole x 1 = 0.20 moles 180. g 2.0 L 1 L = 0.20 M

  9. Molarity Conversion Factors The units in molarity can be used to write conversion factors.

  10. Learning Check Stomach acid is 0.10 M HCl solution. How many moles of HCl are present in 1500 mL of stomach acid? 1) 15 moles HCl 2) 1.5 moles HCl 3) 0.15 mole HCl

  11. Mass Percent The mass percent (%m/m) • Is the g of solute in 100 g of solution. • mass percent = g of solute 100 g of solution • mass percent = g of solute x 100% g of solution

  12. Make 50.0 g 16% (m/m) KCl Solution • Mass percent (%m/m) is calculated from the grams of solute (g KCl) and the grams of solution (g KCl solution). g of KCl = 8.0 g g of solvent (water) = 42.0 g g of KCl solution = 50.0 g 8.0 g KCl (solute) x 100 = 16% (m/m) KCl 50.0 g KCl solution

  13. Learning Check A solution is prepared by mixing 15 g Na2CO3 and 235 g of H2O. Calculate the mass percent (%m/m) of the solution. 1) 15% (m/m) Na2CO3 2) 6.4% (m/m) Na2CO3 3) 6.0% (m/m) Na2CO3

  14. Solution 3) 6.0% (m/m) Na2CO3 mass solute = 15 g Na2CO3 mass solution = 15 g + 235 g = 250 g mass %(m/m) = 15 g Na2CO3 x 100 250 g solution = 6.0% Na2CO3 solution

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