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AP Chemistry Notes Solutions & Colligative Properties.

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Presentation on theme: "AP Chemistry Notes Solutions & Colligative Properties."— Presentation transcript:

1 AP Chemistry Notes Solutions & Colligative Properties

2 A solution is a homogenous mixture of two or more substances. The solute is(are) the substance(s) present in the smaller amount(s). The solvent is the substance present in the larger amount. Solutions…

3 An electrolyte is a substance that, when dissolved in water, results in a solution that can conduct electricity. A nonelectrolyte is a substance that, when dissolved, results in a solution that does not conduct electricity. nonelectrolyte weak electrolyte strong electrolyte

4 Solutions… A saturated solution contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature. An unsaturated solution contains less solute than the solvent has the capacity to dissolve at a specific temperature. A supersaturated solution contains more solute than is present in a saturated solution at a specific temperature.

5 Solutions… Three types of interactions in the solution process: solvent-solvent; solute-solute; and solvent-solute H soln = H 1 + H 2 + H 3

6 Solutions… Two substances with similar intermolecular forces are likely to be soluble in each other. LIKE DISSOLVES LIKE Non-polar molecules are soluble in non-polar solvents CCl 4 in C 6 H 6 Polar molecules are soluble in polar solvents C 2 H 5 OH in H 2 O Ionic compounds are more soluble in polar solvents NaCl in H 2 O or NH 3 (l)

7 Concentration… The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. Percent by Mass: = x 100% mass of solute mass of solution

8 Concentration… Mole Fraction (X) X A = moles of A sum of moles of all components

9 Concentration… Molarity: Molality: M = moles of solute liters of solution m = moles of solute mass of solvent (kg)

10 Concentration… What is the molality of a 5.86 M ethanol (C 2 H 5 OH) solution whose density is 0.927 g/mL? m =m = moles of solute mass of solvent (kg) M = moles of solute liters of solution Assume 1 L of solution: 5.86 moles ethanol = 270 g ethanol 927 g of solution (1000 mL x 0.927 g/mL) mass of solvent = mass of solution – mass of solute = 927 g – 270 g = 657 g = 0.657 kg m =m = moles of solute mass of solvent (kg) = 5.86 moles C 2 H 5 OH 0.657 kg solvent = 8.92 m

11 Solubility… Solid solubility and temperature

12 Solubility… Gas solubility and temperature

13 Solubility… Pressure and Solubility of Gases The solubility of a gas in a liquid is proportional to the pressure of the gas over the solution (Henrys law). Mathematically expressed: c = kP c is the concentration (M) of the dissolved gas P is the pressure of the gas over the solution k is a constant (mol/Latm) that depends only on temperature

14 Henrys Law: low P low c high P high c The solubility of a gas in a liquid is proportional to the pressure of the gas over the solution (Henrys law).

15 Colligative Properties NON-Electrolytic Solutions

16 Colligative Properties… Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. Vapor-Pressure Lowering Boiling Point Elevation Freezing Point Depression Osmosis

17 Vapor Pressure Lowering… Raoults law: P 1 = X 1 P 1 0 P 1 0 = vapor pressure of pure solvent X 1 = mole fraction of the solvent If the solution contains only one solute: X 1 = 1 – X 2 P 1 0 - P 1 = P = X 2 P 1 0 X 2 = mole fraction of the solute

18 Vapor Pressure Lowering… At a given temperature water has a vapor pressure of 22.80 mmHg. Calculate the vapor pressure above a solution of 90.40 g of sucrose (C 12 H 22 O 11 ) in 350.0 mL of water, assuming the water to have a density of 1.000 g/mL.

19 Vapor Pressure Lowering… 90.40 g C 12 H 22 O 11 0.26 mol 350.0 ml H 2 O 19.4 mol X = (19.4/(19.4+0.26)) = 0.9868 V P = XP V P = (0.9868)(22.80) V P = 22.50 mmHg

20 Vapor Pressure Lowering… 23.00 g of an unknown substance was added to 120.0 g of water. The vapor pressure above the solution was found to be 21.34 mmHg. Given that the vapor pressure of pure water at this temperature is 22.96 mmHg, calculate the Molar Mass of the unknown.

21 Vapor Pressure Lowering… 23.0 g X 120.0 g H 2 O 6.65 mol H 2 O VP = 21.34 P = 22.96 V P = XP 21.34 = X (22.96) X = 0.929 X = mol H 2 O / total mol 0.929 = 6.65/(6.65 + x) X mol = 0.506 M = g/mol = 23.0 / 0.506 M = 45.46 g/mol

22 Vapor Pressure Lowering… P B = X B P B 0 P T = P A + P B P T = X A P A 0 + X B P B 0 P A = X A P A 0 Ideal Solution

23 Vapor Pressure Lowering… At 20.0 o C the vapor pressures of methanol (CH 3 OH)and ethanol (C 2 H 5 OH) are 95.0 and 45.0 mmHg respectively. An ideal solution contains 16.1 g of methanol and 92.1 g of ethanol. Calculate the vapor pressure.

24 Vapor Pressure Lowering… 16.1 g CH 3 OH 0.5 mol 92.1 g C 2 H 5 OH 2 mol Total mol = 2.5 V P = X a P a + X b P b V P = (0.5/2.5)(95) + (2/2.5)(45) V P = 55 mmHg

25 Boiling Point Elevation… T b = T b – T b 0 T b is the boiling point of the pure solvent 0 T b is the boiling point of the solution T b > T b 0 T b > 0 T b = K b m m is the molality of the solution K b is the molal boiling-point elevation constant ( 0 C/m)

26 Freezing Point Depression… T f = T f – T f 0 T f > T f 0 T f > 0 T f is the freezing point of the pure solvent 0 T f is the freezing point of the solution T f = K f m m is the molality of the solution K f is the molal freezing-point depression constant ( 0 C/m)

27 Constants…

28 What is the freezing point of a solution containing 478g of ethylene glycol (antifreeze) in 3202 g of water? GIVEN: The molar mass of ethylene glycol is 62.01 g. T f = K f m m =m = moles of solute mass of solvent (kg) = 2.41 m = 3.202 kg solvent 478 g x 1 mol 62.01 g K f water = 1.86 0 C/m T f = K f m = 1.86 0 C/m x 2.41 m = 4.48 0 C T f = T f – T f 0 0 = 0.00 0 C – 4.48 0 C = -4.48 0 C

29 Summary… Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. Vapor-Pressure Lowering P 1 = X 1 P 1 0 Boiling-Point Elevation T b = K b m Freezing-Point Depression T f = K f m Osmotic Pressure ( ) = MRT

30 Colligative Properties Electrolytic Solutions

31 Colligative Properties… Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. 0.1 m NaCl solution 0.1 m Na + ions & 0.1 m Cl - ions 0.1 m NaCl solution0.2 m ions in solution vant Hoff factor (i) = actual number of particles in soln after dissociation number of formula units initially dissolved in soln i should be 123123 Nonelectrolytes NaCl CaCl 2

32 Colligative Properties… Boiling-Point Elevation T b = i K b m Freezing-Point Depression T f = i K f m Osmotic Pressure ( ) = iMRT

33 Colligative Properties… At what temperature will a 5.4 molal solution of NaCl freeze? Solution: T FP = K f m i T FP = (1.86 o C/molal) 5.4 m 2 T FP = (1.86 o C/molal) 5.4 m 2 T FP = 20.1 o C T FP = 20.1 o C FP = 0 – 20.1 = -20.1 o C FP = 0 – 20.1 = -20.1 o C

34 Colligative Properties… Osmotic Pressure ( ): Osmosis is the selective passage of solvent molecules through a porous membrane from a dilute solution to a more concentrated one. A semipermeable membrane allows the passage of solvent molecules but blocks the passage of solute molecules. Osmotic pressure ( ) is the pressure required to stop osmosis. dilute more concentrated

35 Colligative Properties… High P Low P = MRT M is the molarity of the solution R is the gas constant T is the temperature (in K)

36 Colloids… A colloid is a dispersion of particles of one substance throughout a dispersing medium of another substance. Colloid versus solution: Collodial particles are much larger than solute molecules Collodial suspension is not as homogeneous as a solution


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