Test Review ep 5 text only

Test Review EP 5 Sept 2010
Gary Sakuma
Like dissolves like
What is a polar molecule?
Water
Dissolves ionics, sugars, not lipids, methane, hydrocarbons
Formulae and percent
calcium chloride dihydrate
Percent by water
Solubility is affected by
Temperature
Pressure
size
Mole fraction problem
Calculate the mole fraction of each component in a solution of 1.50 mol ethanoic acid (CH3COOH) in 12.00 mol of water.
Colligative properties
What is the freezing point of a solution of 20.0 g of CCl4 dissolved in 500.0 g of benzene? The freezing point of benzene is 5.48°C; Kf is 5.12°C/m.
m/m problem
Calculate the grams of solute required to make the following solutions.
1100 g of saline solution (0.90% NaCl (m/m))
molarity
How many grams of NaCl would you need to make 500 ml of 3.0M

Concentration of Solute
The amount of solute in a solution is given by its concentration.
Molality
What is the molality of 5 moles of KCl in 500 ml of water
Two Other Concentration Units
V/V problems
What is the concentration (in % (v/v)) of the following solutions?
55 mL of ethanol (C2H5OH) is diluted to a volume of 250 mL with water.
Calculating Concentrations
Calculate molality






Boiling Point Elevation and Freezing Point Depression
∆T = K•m•i
i = van’t Hoff factor = number of particles produced per molecule/formula unit. For covalent compounds, i = 1. For ionic compounds, i = the number of ions present (both + and -)
Compound Theoretical Value of i
glycol 1
NaCl 2
CaCl2 3
Ca3(PO4)2 5
Boiling Point Elevation and Freezing Point Depression
∆T = K•m•i
Change in Boiling Point
Dissolve 62.1 g of glycol (1.00 mol) in 250. g of water. What is the boiling point of the solution?
Kb = 0.52 oC/molal for water (see Kb table).
Solution ∆TBP = Kb • m • i

1. Calculate solution molality = 4.00 m
2. ∆TBP = Kb • m • i
∆TBP = 0.52 oC/molal (4.00 molal) (1)
∆TBP = 2.08 oC
BP = 100 + 2.08 = 102.08 oC (water normally boils at 100)

Freezing Point Depression
Calculate the Freezing Point of a 4.00 molal glycol/water solution.
Kf = 1.86 oC/molal (See Kf table)
Solution
∆TFP = Kf • m • i
= (1.86 oC/molal)(4.00 m)(1)
∆TFP = 7.44
FP = 0 – 7.44 = -7.44 oC (because water normally freezes at 0)


Freezing Point Depression
At what temperature will a 5.4 molal solution of NaCl freeze?
Solution
∆TFP = Kf • m • i
∆TFP = (1.86 oC/molal) • 5.4 m • 2
∆TFP = 20.1 oC
FP = 0 – 20.1 = -20.1 oC
Solution Molality
How many grams of potassium iodide must be dissolved in 500.0 grams of water to produce a 0.060 molal KI solution
.5 kg H20 * .6 mole KI/1 kg H2O *
166.0 gKI/1 mol KI
= answer
Types of problems on exam
Ch 15 and 16 vocabulary and reading
Mass/Mass m/m
Volume/volume v/v
Hydrates percent water and writing formulae
Molarity
Molality
Freezing point depression/boiling point elevation T=k m I