 # Solution Concentration. Review  A solution is a homogeneous mixture.  The solvent is the major component of the solution.  The solute is the minor.

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Solution Concentration

Review  A solution is a homogeneous mixture.  The solvent is the major component of the solution.  The solute is the minor component and active ingredient.  A saturated solution holds the maximum amount of solute that is theoretically possible for a given temperature.

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Solution Concentration  Is one glass of tea stronger than the other? –What’s true about the “stronger” glass of tea? –How much tea does it have in it compared to the other glass?

Solution Concentration  Concentration – a ratio comparing the amount of solute to the amount of solution.  Many ways of expressing concentration: –% by weight (% w/w) –% by volume (% v/v) –parts per million (ppm) or parts per billion (ppb) for very low concentrations –molality (m) –Molarity (M)

Concentrated vs. Dilute  The words “concentrated” and “dilute” are opposites.  EX: The dark tea is more concentrated than the light tea.  EX: The light tea is more dilute than the dark tea.

Concentrated vs. Dilute Concentrated solution Dilute solution = solute particles

Percent by Weight  % by weight (% w/w)  What is the % w/w of a solution if 3.00 grams of NaCl are dissolved in 17.00 g of water? –mass of solute = 3.00 g –mass of solution = 3.00 g + 17.00 g = 20.00 g –(3.00 g / 20.00 g) x 100% = 15.0% w/w

Percent by Volume  % by volume (% v/v)  What is the % v/v of a solution if 20.0 mL of alcohol are dissolved in 50.0 mL of solution? –volume of solute = 20.0 mL –volume of solution = 50.0 mL –(20.0 mL / 50.0 mL) x 100% = 40.0%

Molarity  Molarity (M) –UNITS: mol/L or Molar (M) –Example: 0.500 mol/L = 0.500 M

Molarity  What is the Molar concentration of a sol’n if 20.0 grams of KNO 3 (MM = 101.11 g/mol) is dissolved in enough water to make 800. mL? –Convert g of KNO 3 to mol of KNO 3 –Convert mL to L

Molarity  What is the Molar concentration of a sol’n if 0.198 mol KNO 3 is dissolved in enough water to make 0.800 L?

Molarity  What is the Molar concentration of a sol’n if 10.5 grams of glucose (MM = 180.18 g/mol) is dissolved in enough water to make 20.0 mL of sol’n? –Convert g of glucose to mol of glucose. –Convert mL to L.

Molarity  What is the Molar concentration of a sol’n if 0.0583 mol of glucose is dissolved in enough water to make 0.0200 L of sol’n?

Calculating Grams  How many grams of KI (MM = 166.00 g/mol) are needed to prepare 25.0 mL of a 0.750 M solution? –Convert mL to L. –Solve for moles. –moles of KI = 0.750 M x 0.0250 L = 0.0188 mol KI

Calculating Grams  How many grams of KI (MM = 166.00 g/mol) are needed to prepare 25.0 mL of a 0.750 M solution? –Convert 0.0188 mol KI to grams.

Calculating Grams  How many grams of HNO 3 (MM = 63.02 g/mol) are present in 50.0 mL of a 1.50 M sol’n? –Convert mL to L. 50.0 mL = 0.0500 L –Solve for moles: moles = (1.50 M)(0.0500 L) = 0.0750 mol HNO 3 –Convert 0.0750 mol HNO 3 to grams: 0.0750 mol HNO 3 = 4.73 g HNO 3

Dilution  Dilute (verb) - to add solvent to a solution. –Decreases sol'n concentration. –M 1 V 1 = M 2 V 2 M 1 = initial conc. V 1 = initial volume M 2 = final conc. V 2 = final volume –Assumes no solute is added.

Dilution Stock Solution Impractically High Concentration Usable Solution Add H 2 O Question for Consideration: Why do you think chemical supply companies typically sell acids (and other solutions) in extremely high concentrations when it would be safer to ship more dilute solutions?

Dilution  To what volume should 40.0 mL of 18 M H 2 SO 4 be diluted if a concentration of 3.0 M is desired? –What do we want to know? V 2 –What do we already know? M 1 = 18 M V 1 = 40.0 mL M 2 = 3.0 M –(18 M)(40.0 mL) = (3.0 M)V 2 –720 M*mL = (3.0 M)V 2 –V 2 = 240 mL

Dilution  You are asked to prepare 500. mL of 0.250 M HCl, starting with a 12.0 Molar stock sol'n. How much stock should you use? –What do we want to know? V 1 –What do we already know? M 1 = 12.0 M M 2 = 0.250 M V 2 = 500. mL –(12.0 M) V 1 = (0.250 M)(500. mL)‏ –(12.0 M) V 1 = 125 M*mL –V 1 = 10.4 mL

 To how much water should you add 20.0 mL of 5.00 M HNO 3 to dilute it to 1.00 M? –What do we want to know? How much water to add. (V 2 - V 1 )‏ –What do we already know? M 1 = 5.00 M V 1 = 20.0 mL M 2 = 1.00 M –(5.00 M)(20.0 mL) = (1.00 M) V 2 –100. M*mL = (1.00 M) V 2 –V 2 = 100. mL –Water added = 100. mL - 20.0 mL = 80. mL Dilution

Colligative Properties  Colligative properties are properties of solutions that are affected by the number of particles but not the identity of the solute

Boiling Point Elevation: A Colligative Property  Boiling point elevation is the temperature difference between a solution and pure solvent  The value of the boiling point elevation is directly proportional to molality, meaning the greater the number of solute particles, the greater the elevation

Freezing Point Depression: A Colligative Property  Freezing point depression is the difference in temperature between a solution and a pure solvent  The value of freezing point depression is directly proportional to molality, meaning the greater the number of solute particles

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