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  1. Solutions Lesson 2

  2. Solution Concentration • The concentration of a solution is a measure of how much solute is dissolved in a specific amount of solvent or solution. • Concentration may be described qualitatively using the words concentrated or dilute. • A concentrated solution contains a large amount of solute. • A dilute solution contains a small amount of solute.

  3. Expressing Concentration • Some commonly used quantitative descriptions of a solution’s concentration are • Percent by mass: (mass of solute/mass of solution) x 100% • Percent by volume: (volume of solute/volume of solution) x 100% • Molarity: moles of solute/liter of solution • Molality: moles of solute/kg of solvent • Mole fraction: moles of solute/(moles of solute + moles of solvent)

  4. Example Problem Percent by Mass In order to maintain a sodium chloride concentration similar to ocean water, an aquarium must contain 3.6 g NaCL per 100.0 g of water. What is the percent by mass of NaCl in the solution? Do Practice Problems 8-10 on page 463.

  5. Example Problem: Molarity Molarity (M) is the number of moles of a solute dissolved per liter of solution. It is also called the molar concentration. A 100.5 mL intravenous solution contains 5.10 g of glucose (C6H12O6). The molar mass of glucose is 180.16 g/mol. What is the molarity of this solution? Do practice problems 14-16 page 465

  6. Preparing Molar Solutions • Step 1: The mass of the solute to be used is measured. • Step 2: The solute is placed in a volumetric flask of the correct volume. • Step 3: Distilled water is added to the flask to bring the solution level up to the calibration on the flask.

  7. Example Problem: Preparing Molar Solutions How many grams of CaCl2 would be dissolved in 1.0 L of a 0.10 M solution of CaCl2? 1 mole CaCl2 = 110.98 g 1.0 M = 110.98 g CaCl2/1.0 L Multiply both sides by 0.1 0.10 M = 11.098 g CaCl2 Do practice problems 18-20 on page 466.

  8. Diluting Solutions • In order to dilute a concentrated solution, more solvent must be added. • The number of moles of solute does not change. • Only the volume of the solution changes. • This relationship can be written as M1V1 = M2V2.

  9. Example Problem: Diluting Stock Solutions What volume, in milliliters, of 2.00 M calcium chloride (CaCl2 ) stock solution would you use to make 0.50 L of 0.300 M calcium chloride solution? M1 = 2.00 V1 = ? M2 = 0.300 V2 = 500 mL 2.00 (V1) = 0.300 (500) V1 = Do practice problems 21 – 23 on page 469.

  10. Molality • The molality (m) of a solution is the number of moles of a solute dissolved in 1kg of solvent. • A 1 molal solution has 1 mole of solute dissolved in 1000 g of solvent. • Example problem A student adds 4.5 g NaCl to 100.0 g of water. Calculate the molality of the solution. Mass = 4.5 g Molar Mass NaCl = 58.45 g/mol MolNaCl = 4.5g/58.45 g/mol = 0.077 molNaCl Mass of water = 100 g; convert to kg 100.0 g/1000 g = 0.1 kg 0.077 molNaCl / 0.1 kg = 0.77 molality

  11. Mole Fraction The mole fraction of a solute to solution is the number of moles of solute divided by the total number of moles of solute and solvent. Mole fractions can be calculated for both the solute and the solution Xa = na/na +nb Xa is the mole fraction, na is the moles of solute, nb is the moles of solvent

  12. Example Problem A solution is composed of 62.5% water and 37.5% HCl. What are the mole fractions of water and HCl in solution? Assume 100g of solution 62.5% water = 62.5 g H2O 37.5% HCl = 37.5g HCl 62.5g/18.02g/mol = 3.47 mol H2O 37.5 g/36.46g/mol = 1.03 molHCl XH20 = 3.47/3.47+1.03 = 0.77 XHCl = 1.03/4.5 = 0.23 Do practice problem 26 and 27 on page 470.