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II. Solution Concentration (p. 480 – 488)

Ch. 14 – Mixtures & Solutions. II. Solution Concentration (p. 480 – 488). A. Concentration. The amount of solute in a solution Describing Concentration % by mass - medicated creams % by volume - rubbing alcohol ppm, ppb - water contaminants molarity - used by chemists

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II. Solution Concentration (p. 480 – 488)

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  1. Ch. 14 – Mixtures & Solutions II. Solution Concentration(p. 480 – 488)

  2. A. Concentration • The amount of solute in a solution • Describing Concentration • % by mass - medicated creams • % by volume - rubbing alcohol • ppm, ppb - water contaminants • molarity - used by chemists • molality - used by chemists

  3. substance being dissolved total combined volume A. Molarity • Concentration of a solution

  4. A. Molarity 2M HCl What does this mean?

  5. B. Molarity Calculations LITERS OF GAS AT STP Molar Volume (22.4 L/mol) MASS IN GRAMS NUMBER OF PARTICLES MOLES Molar Mass (g/mol) 6.02  1023 particles/mol Molarity (mol/L) LITERS OF SOLUTION

  6. B. Molarity Calculations • How many grams of NaCl are required to make 0.500L of 0.25M NaCl? 58.44 g NaCl 1 mol NaCl 0.500 L sol’n 0.25 mol NaCl 1 L sol’n = 7.3 g NaCl

  7. B. Molarity Calculations • Find the molarity of a 250 mL solution containing 10.0 g of NaF. 10.0 g NaF 1 mol NaF 41.99 g NaF = 0.95 M NaF .25 L sol’n

  8. C. Dilution • Preparation of a desired solution by adding water to a concentrate. • Moles of solute remain the same.

  9. C. Dilution • What volume of 15.8M HNO3 is required to make 250 mL of a 6.0M solution? GIVEN: M1 = 15.8M V1 = ? M2 = 6.0M V2 = 250 mL WORK: M1 V1 = M2 V2 (15.8M)V1 = (6.0M)(250mL) V1 = 95 mL of 15.8M HNO3

  10. mass of solvent only 1 kg water = 1 L water D. Molality

  11. D. Molality • Find the molality of a solution containing 75 g of MgCl2 in 250 mL of water. 75 g MgCl2 1 mol MgCl2 95.21 g MgCl2 0.25 kg water = 3.2m MgCl2

  12. D. Molality • How many grams of NaCl are req’d to make a 1.54m solution using 0.500 kg of water? 0.500 kg water 1.54 mol NaCl 1 kg water 58.44 g NaCl 1 mol NaCl = 45.0 g NaCl

  13. 500 mL of 1.54M NaCl 500 mLwater 500 mL volumetric flask 500 mL mark 45.0 gNaCl E. Preparing Solutions • 1.54m NaCl in 0.500 kg of water • mass 45.0 g of NaCl • add water until total volume is 500 mL • mass 45.0 g of NaCl • add 0.500 kg of water

  14. 95 mL of15.8M HNO3 250 mL mark water for safety E. Preparing Solutions • 250 mL of 6.0M HNO3by dilution • measure 95 mL of 15.8M HNO3 • combine with water until total volume is 250 mL • Safety: “Do as you oughtta, add the acid to the watta!” or AA – add acid!

  15. Turn in one paper per team. Complete the following steps: A) Show the necessary calculations. B) Write out directions for preparing the solution. C) Prepare the solution. For each of the following solutions: 1) 100.0 mL of 0.50M NaCl 2) 0.25m NaCl in 100.0 mL of water 3) 100.0 mL of 3.0M HCl from 12.0M concentrate. (don’t actually prepare this one!) Solution Preparation Mini-Lab

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