# 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

WHICH, VAPOR PRESSURE = EXTERNAL ATMOSPHERIC PRESSURE Colligative Properties: Boiling Point Boiling-Point Elevation

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.

Figure 12.11 Figure 12.11a Figure 12.11b Colligative Properties

Osmosis Osmotic pressure, , is the pressure required to stop osmosis: V nRT n RT

V MRT DETERMINATION OF MOLAR MASS Usually use freezing point depression or boiling point elevation.

Process Calculate molality or Molarity Need to be given mass of solute. Calculate molar mass: mole = grams A 1.2436 gram sample of an unknown electrolyte is dissolved in 10.9303 grams of

benzophenone which produces a solution that freezes as 43.2 C. If pure benzophenone melts at 48.1 C, what is the molecular weight of the unknown compound? Kf =9.8 C/m Tf = Kf m

Tf = 4.9 C 4.9 C = [9.8 C/m] [m] [m] = 0.50 moles in one kg of benzophenone Used 0.010903 kg of solvent. [0.50 moles/kg] [0.010903 kg] = 0.0054515 moles

Molar mass = grams of compound/moles of compound Molar Mass = 10.903 grams/0.50 moles = 21.81 grams/mole Colligative Properties of

Electrolyte Solutions Electrolytes: Solutes which dissociate into ions in solution Ionic Solutes (NaCl) Acids (HCl) Bases

Colligative Properties of Electrolyte Solutions I = actual number of particles in solution after dissociation/number of formula units initially dissolved in soln Tf = i Kb m

Tb = i Kf m = i MRT Problem 12.72: Page 549 Arrange the following aqueous solutions in order of decreasing freezing point, and explain your reasoning. 0.050 HCl; 0.50 m

glucose; 0.50 m acetic acid.

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