1. Solutions

2.  Colligative Property  property that depends on the concentration of solute particles, not their identity

3.  Freezing Point Depression  Freezing Point Depression (  t f )  f.p. of a solution is lower than f.p. of the pure solvent  Boiling Point Elevation  Boiling Point Elevation (  t b )  b.p. of a solution is higher than b.p. of the pure solvent

4. View Flash animation.Flash animation Freezing Point Depression

5. Solute particles weaken IMF in the solvent. Boiling Point Elevation

6.  Applications  salting icy roads  making ice cream  antifreeze  cars (-64°C to 136°C)  fish & insects

7.  t :change in temperature (° C ) k :constant based on the solvent (° C·kg/mol ) m :molality ( m ) n :# of particles  t = k · m · n

8.  # of Particles  Nonelectrolytes (covalent)  remain intact when dissolved  1 particle  Electrolytes (ionic)  dissociate into ions when dissolved  2 or more particles

9.  At what temperature will a solution that is composed of 0.73 moles of glucose in 225 g of phenol boil? m = 3.2m n = 1  t b = k b · m · n WORK: m = 0.73mol ÷ 0.225kg GIVEN: b.p. = ?  t b = ? k b = 3.60°C·kg/mol  t b = (3.60°C·kg/mol)(3.2m)(1)  t b = 12°C b.p. = 181.8°C + 12°C b.p. = 194°C

10.  Find the freezing point of a saturated solution of NaCl containing 28 g NaCl in 100. mL water. m = 4.8m n = 2  t f = k f · m · n WORK: m = 0.48mol ÷ 0.100kg GIVEN: f.p. = ?  t f = ? k f = 1.86°C·kg/mol  t f = (1.86°C·kg/mol)(4.8m)(2)  t f = 18°C f.p. = 0.00°C - 18°C f.p. = -18°C

11. Percent Solutions  If both solute & solvent are liquids Percent by volume (% v/v) = volume of solute × 100% solution volume  If a solid is dissolved in a liquid Percent (mass/volume) (%(m/v)) = mass of solute (g) × 100% solution volume (mL) Must be the same unit: mL or L Must be this unit

12. Example 1 What is the percent by volume of ethanol (C 2 H 6 O) or ethyl alcohol, in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water? Volume of solute = 85 mL Volume of solution = 250 mL % (v/v) = 85 mL ethanol × 100% 250 mL solution = 34% ethanol % (v/v) = volume of solute × 100% volume of solution

13. Example 2 How many grams of glucose (C 6 H 12 O 6 ) would you need to prepare 2.0 L of 2.8% glucose (m/v) solution? Solution volume = 2.0 L → change to mL Percent by mass = 2.8% Percent (mass/volume) (%(m/v) = mass of solute (g) × 100% solution volume (mL) 2.8% = mass of solute (g) × 100% 2,000 mL 2.0L 100% 0.028 = X 2,000 mL X = 56 g of solute 1L 1000mL= 2,000 mL

14. 1. What is the concentration, in percent (m/v), of a solution with 75g K 2 SO 4 in 1500mL of solution? 2. A bottle of hydrogen peroxide antiseptic is labeled 3.0% (v/v). How many mL H 2 O 2 are in a 400.0 mL bottle of this solution? 3. Calculate the grams of solute required to make 250 mL of 0.10% MgSO 4 (m/v). Percent Solution Problems You do not have to write the problem. You MUST show your work.