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Chapter16 Solutions

Chapter16 Solutions. 16.2 Concentrations of Solutions. Chemistry. Today we are learning to:- 1. Use various ways to describe the concentration of a solution 2. Solve problems involving molarity , percent by volume , percent by mass and parts per million of a solution. Molarity.

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Chapter16 Solutions

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  1. Chapter16 Solutions 16.2 Concentrations of Solutions

  2. Chemistry Today we are learning to:-1. Use various ways to describe the concentration of a solution 2. Solve problems involving molarity, percent by volume, percent by mass and parts per million of a solution

  3. Molarity • The concentration of a solution is the amount of solute dissolved in a given quantity of solvent. • A dilute solution contains a small amount of solute. • Aconcentrated solution contains a large amount of solute. • These terms are qualitative only. Molarity expresses the concentration of a solution quantitatively or mathematically.

  4. Molarity • Molarity (M) = moles of solute dissolved in one liter of solution. • The unit of measurement for Molarity(M) is moles per liter • 1mole of solute dissolved in 1liter of solvent has a concentration of 1mol/L also called 1Molar or 1M

  5. Exothermic and Endothermic Processes 17.1 Molarity Ex.1. What is the molarity of a solution that contains 4.0moles of NaOH in 0.5L of solution? 1. Identify the knowns and unknowns Amount of NaOH = 4.0mol Volume of solution = 0.5L Molarity = ? Molarity (M) = 4.0mol = 8.0M 0.5L

  6. Exothermic and Endothermic Processes 17.1 Molarity Ex.1. What is the molarity of a compound that contains 82.0g of Ca(NO3)2 in 2.0L of solution? Identify the knowns and unknowns Amount of Ca(NO3)2 = 82.0g Volume of solution = 2.0L Molarity = ? Step 3 Calculate the molarity Molarity (M) = moles = 0.500mol = 0.250M liters 2.0L Step 1 Calculate the formula mass of Ca(NO3)2 Ca = 1x40.g = 40.g N = 2x14g = 28 g O = 6x16.g= 96.g__ Formula mass = 164g/mol Step 2 Convert mass of Ca(NO3)2 to moles = 82.0g = 0.500mol 164g/mol moles = grams x ----------------------- 1 mol Formula mass

  7. Molarity • Try review questions 24-29 on page 128. Show all working out

  8. Percent Solutions • The concentration of a solution in percent can be expressed in two ways: • ratio of the volume of the solute to the volume of the solution (liquids in liquids) • ratio of the mass of the solute to the mass of the solution (solids in solids)

  9. Exothermic and Endothermic Processes 17.1 Percent Solutions Ex.1. What is the Percent by volume of ethanol (CH3CH2OH) in the final solution if 85 mL of ethanol is diluted to 250 mL with water? 1. Identify the knowns and unknowns Volume of ethanol = 85mL Volume of solution = 250L % ethanol (v/v) = ? % ethanol (v/v) = 85mL x 100%= 34% ethanol 250L

  10. Exothermic and Endothermic Processes 17.1 Percent Solutions Ex.1. What is the Percent by mass of sodium hydroxide(NaOH) if 2.50g of sodium hydroxide are dissolved in 50.00g of water? 1. Identify the knowns and unknowns Mass od NaOH = 2.50g Mass of water = 50.00g % NaOH (m/m) = ? % NaOH(m/m) = 2.50g x 100%= 4.76% NaOH 52.50g

  11. Parts per Million • This measurement of concentration is useful for extremely dilute solutions. • Calculations are similar to % calculations except multiply by 1,000,000 instead of 100

  12. Exothermic and Endothermic Processes 17.1 Parts per Million Ex.1. The solubility of AgCl is 0.008 grams/100 grams of water. What is this concentration in ppm? Concentration = 0.008gX 1,000,000 = 80 ppm 100g Ex.2. A certain pesticide has a toxic solubility of 5.0 grams/Kg of body weight. What is this solubility in ppm? First convert Kg to g: 1Kg = 1000g Concentration = __5g__X 1,000,000 = 5000 ppm 1000g

  13. 17.1 Parts per Million • Ex.3. A 2.0 Kg sample of water is found to contain 6.7 x 10 -3 g Pb. Express this concentration in ppm. First convert Kg to g: 1Kg = 1000g Concentration = 6.7 x 10 -3gX 1,000,000 2000g = 3.35 ppm • Try review questions 30-33 on page 128. Show all working out

  14. END OF SHOW

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