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Colligative Properties of Solutions

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1 Colligative Properties of Solutions
Chemistry XII Chapter I ZONALIA FITRIZA Colligative Properties of Solutions To play the movies and simulations included, view the presentation in Slide Show Mode.

2 SOLUTION Solution is homogeneous mixture of solute and solvent
SOLUTE – the part of a solution that is being dissolved (usually the lesser amount) SOLVENT – the part of a solution that dissolves the solute (usually the greater amount) Solute + Solvent = Solution

3

4 = Molarity MOLARITY MOLARITY, MOLALITY AND MOLE FRACTION ( M )
Molarity is the concentration unit that express mole of solute in 1 L of solvent Molarity ( M ) = moles solute liters of solution

5 moles = M•V USING MOLARITY What mass of oxalic acid, H2C2O4, is
required to make 250. mL of a M solution? moles = M•V Step 1: Change mL to L. 250 mL * 1L/1000mL = L Step 2: Calculate. Moles = ( mol/L) (0.250 L) = moles Step 3: Convert moles to grams. ( mol)(90.00 g/mol) = g

6 Concentration Units An IDEAL SOLUTION is one where the properties depend only on the concentration of solute. Need conc. units to tell us the number of solute particles per solvent particle. The unit “molarity” does not do this!

7 MOLALITY MOLALITY, m m of solution = mol solute kilograms solvent
Molality is the concentration unit that shows the mole of solute in 1 kg of solvent MOLALITY, m m of solution = mol solute kilograms solvent % by mass grams solute grams solution % by mass =

8 Calculating Concentrations
Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H2O. Calculate molality and % by mass of ethylene glycol.

9 Calculating Concentrations
Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H2O. Calculate m & % of ethylene glycol (by mass). Calculate molality Calculate weight %

10 Try this molality problem
25.0 g of NaCl is dissolved in mL of water. Find the molality (m) of the resulting solution. m = mol solute / kg solvent 25 g NaCl mol NaCl 58.5 g NaCl = mol NaCl Since the density of water is 1 g/mL, 5000 mL = 5000 g, which is 5 kg 0.427 mol NaCl 5 kg water = m salt water

11 Mole Fraction of solute Mole Fraction of solute
MOLE FRACTION (X) Mole fraction is the number that shows the ratio of moles of solute or solvent with the mole of solution (solute +solvent) Mole Fraction of solute Mole Fraction of solute XA = nA nA + nB XB = nB nA + nB XA = mole fraction of solute XB = mole fraction of solvent nA = mole of solute nB = mole of solvent

12 LEARNING CHECK Calculate mole fraction of sugar (C12H22O11) in any solution obtained by dissolving 17,1 g of sugar in 1 kg of water (H2O)! Ar C = 12, H =1 O = 16 2. Calculate mole fraction of solution that containing 18% of glucose (Mr = 180)!

13 Colligative Properties
On adding a solute to a solvent, the properties of the solvent are modified. Vapor pressure decreases Melting point decreases Boiling point increases Osmosis is possible (osmotic pressure) These changes are called COLLIGATIVE PROPERTIES. They depend only on the NUMBER of solute particles relative to solvent particles, not on the KIND of solute particles.

14 Colligative Properties of Nonelectrolyte Solutions
Colligative properties are properties that depend only on the number of solute particles in solution and not on the kinds of the solute particles. Vapor-Pressure Lowering/Depression Rault’s Law state that saturated vapor pressure of a component of solution that can evaporate is equal with the saturated vapor pressure of pure solvent times the mole fraction in certain temperature. P = vapor pressure of pure solvent P = XB P XB = mole fraction of the solvent Raoult’s law 12.6

15 X2 = mole fraction of the solute
Vapor pressure depression (P) is the difference of saturated vapor pressure of solvent with saturated vapor pressure of solution XA = 1 – XB P - P = DP X2 = mole fraction of the solute DP = XA Po

16 Boiling-Point Elevation
DTb = Tb – T b T b is the boiling point of the pure solvent T b is the boiling point of the solution Tb > T b DTb > 0 DTb = Kb m m is the molality of the solution Kb is the molal boiling-point elevation constant (0C/m) 12.6

17 Fractional Distillation Apparatus
12.6

18 Change in Boiling Point
Common Applications of Boiling Point Elevation

19 Freezing-Point Depression
DTf = T f – Tf T f is the freezing point of the pure solvent T f is the freezing point of the solution T f > Tf DTf > 0 DTf = Kf m m is the molality of the solution Kf is the molal freezing-point depression constant (0C/m) 12.6

20 12.6

21 DTf = Kf m Kf water = 1.86 0C/m DTf = Kf m
What is the freezing point of a solution containing 478 g of ethylene glycol (antifreeze) in 3202 g of water? The molar mass of ethylene glycol is g. DTf = Kf m Kf water = C/m = 3.202 kg solvent 478 g x 1 mol 62.01 g m = moles of solute mass of solvent (kg) = 2.41 m DTf = Kf m = C/m x 2.41 m = C DTf = T f – Tf Tf = T f – DTf = C – C = C 12.6

22 Change in Freezing Point
Which would you use for the streets of Bloomington to lower the freezing point of ice and why? Would the temperature make any difference in your decision? sand, SiO2 Rock salt, NaCl Ice Melt, CaCl2

23 Change in Freezing Point
Ethylene glycol/water solution Pure water The freezing point of a solution is LOWER than that of the pure solvent

24 Change in Freezing Point
Common Applications of Freezing Point Depression Ethylene glycol – deadly to small animals Propylene glycol

25 Change in Freezing Point
Common Applications of Freezing Point Depression Which would you use for the streets of Bloomington to lower the freezing point of ice and why? Would the temperature make any difference in your decision? sand, SiO2 Rock salt, NaCl Ice Melt, CaCl2

26 Osmotic Pressure (p) 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 (p) is the pressure required to stop osmosis. more concentrated dilute 12.6

27 Osmotic Pressure (p) p = MRT 12.6 High P Low P
M is the molarity of the solution R is the gas constant T is the temperature (in K) 12.6

28 A cell in an: isotonic solution hypotonic solution hypertonic solution
12.6

29 Chemistry In Action: Desalination

30 Colligative Properties of Nonelectrolyte Solutions
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 P1 = X1 P 1 Boiling-Point Elevation DTb = Kb m Freezing-Point Depression DTf = Kf m Osmotic Pressure (p) p = MRT 12.6

31 Colligative Properties of Electrolyte Solutions
0.1 m NaCl solution 0.1 m Na+ ions & 0.1 m Cl- ions 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 0.2 m ions in solution van’t Hoff factor (i) = actual number of particles in soln after dissociation number of formula units initially dissolved in soln i should be nonelectrolytes 1 NaCl 2 CaCl2 3 12.7

32 Colligative Properties of Electrolyte Solutions
Boiling-Point Elevation DTb = i Kb m Freezing-Point Depression DTf = i Kf m Osmotic Pressure (p) p = iMRT 12.7


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