CHAPTER 12: Concentration Terms for Solutions Concentration =
CHAPTER 12: Concentration Terms for Solutions Concentration = Amount solute/amount solvent Some Concentration Terms are Temperature Sensitive
Concentration Terms Molarity (M) = moles of solute/liter(s) of solution TEMPERATURE SENSITIVE How many grams of NH3 are in 25.5 ml of
0.15M NH3 (aq) ? CONCENTRATION TERMS: Molality Molality (m) = moles of solute/kg of solvent
Molality (m) Not Temperature Sensitive What is the molal (molality) of glucose C6H12O6, a sugar found in many fruits, in a solution made by dissolving 24.0 g of glucose in 1.0 kg of water.
Concentration Terms: Mole Fraction Mole Fraction (X) = moles component/total moles in solution Not Temperature Sensitive
Ways of Expressing Concentration Mole Fraction, Molarity, and Molality A solution is made containing 7.5 grams CH3OH in 245 grams H2O. Calculate (a) the mole fraction of CH3OH
Concentration Terms: Percent by Mass Percent by Mass= (weight of solute/total weight of solution) (100) Not Temperature Sensitive
A solution is made containing 7.5 grams CH 3OH in 245 grams H2O. Calculate (b) the mass percent of CH3OH Chang Text Problem 12.22: Page 546 The concentrated sulfuric acid we use in the
laboratory is 98.0 percent H2SO4 by mass. Calculate the molality and molarity of the acid solution. The density of the solution is 1.83 g/ml. Chang 12.24: Page 547 The density of an aqueous solution
containing 10 percent of ethanol (C 2H5OH) by mass is 0.984 g/ml. (a) Calculate the molality; (b) Calculate the molarity; (C ) What volume of the solution would contain 0.125 mole of ethanol? Problem 12.22: Page 535
1. Assume you have 100 grams of solution. 2. Mass % = [mass solute/mass solute + mass H2O] 100 3. Have 98 grams of solute (98%). Have 2.0 grams of H2O. 4. Molality = moles of solute/kg of solvent
Problem 12.22: Page 535 Molarity = moles solute/liters of solution Have 100g of solution. Use density of the sulfuric acid solution to calculate volume. Calculate moles of sulfuric in 98.0 grams Colligative Properties
Physical properties of a solvent which depend upon the amount of solute present in a solution. (1) Lowering the Vapor Pressure Non-volatile solvents reduce the ability of the surface solvent molecules to escape the liquid. TEMPERATURE SENSITIVE
The amount of vapor pressure lowering depends on the amount of solute. Colligative Properties Of Nonelectrolyte Solutions: Vapor Pressure Raoults Law Raoults Law: PA is the vapor pressure with solute,
PA is the vapor pressure without solvent, and A is the mole fraction of A, then PA A PA Recall Daltons Law: PA A Ptotal
Raoults Law Remember in a Gas Mixture PT = Pi(one) + Pi(two) + pi(three) With Multiple Solutes Ptotal = XA(PA) + XB(PB) + .
Colligative Properties: Vapor Pressure Raoults Law Ideal solution: one that obeys Raoults law. PA =(X solute) (Vapor Pressure Pure Solvent) Boiling-Point Elevation
Goal: interpret the phase diagram for a solution. Non-volatile solute lowers the vapor pressure. Therefore the triple point - critical point curve is lowered. Problem Calculate the mass of propylene glycol
(C3H8O2) that must be added to 0.500 kg of water to reduce the vapor pressure by 4.60 torr at 40 C. Colligative Properties: Boiling Point BOILING POINT IS TEMPERATURE AT
Non-volatile solute lowers the vapor pressure. Therefore the triple point - critical point curve is lowered. Tb = Kb m
Colligative Properties Colligative Properties: Freezing-Point Depression The solution freezes at a lower temperature (Tf) than the pure solvent.
T f K f m T f K f m Colligative Properties: Freezing Point Freezing-Point Depression When a solution freezes, almost pure solvent is formed first.
Therefore, the sublimation curve for the pure solvent is the same as for the solution. Therefore, the triple point occurs at a lower temperature because of the lower vapor pressure for the solution. The melting-point (freezing-point) curve is a
vertical line from the triple point. Colligative Properties Table 12.2 Using data from table, calculate the
freezing and boiling points of each of the following solutions. (a) 0.35 m glycerol (C3H8O3) in ethanol T b = K b m from table 13.4, normal boiling point for ethanol is 78.4 C; Kb is 1.22 C/m
Tf = Kf m Normal freezing point for ethanol is -114.6 Celsius degrees; Kf 1.99 C/m
Problem 12.56: Page 548 An aqueous solution contains the amino acid glycine (NH2CH2COOH. Assuming that the acid does not ionize in water, calculate the molality of the solution if it freezes at -1.1 Celsius.
T f = Kf m Kf = 1.86 Celsius/m Colligative Properties Freezing-Point Depression
Colligative Properties: Osmosis Osmosis Movement of a solvent from an area of high solvent concentration to an area of low solvent concentration across a semi permeable membrane.
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