Salting The Roads Colligative Property. B. Types. A. Definition

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Salting The Roads Colligative Property A. Definition property that depends on the concentration of solute particles, not their identity Unit 11: Solutions Lesson 11.3: Colligative Properties 68 B.

This has nothing to do with the type of solute. It only has to do with the amount. These properties (that have to do with the amount of a solute, not the type) are called Colligative Properties.

When we dissolve something in water, the water molecules surround the solute particles and weakly bond to them (IMAF s).

1 Salting The Roads Colligative Property A. Definition property that depends on the concentration of solute particles, not their identity Unit 11: Solutions Lesson 11.3: Colligative Properties 68 B. Types Lowering Vapor Pressure Freezing Point Depression ( t f ) f.p. of a solution is lower than f.p. of the pure solvent Boiling Point Elevation ( t b ) b.p. of a solution is higher than b.p. of the pure solvent A Strange Thing Solutions have higher boiling points and lower freezing points than pure samples of their solvents. This has nothing to do with the type of solute. It only has to do with the amount. These properties (that have to do with the amount of a solute, not the type) are called Colligative Properties. It explains: Why antifreeze works. Why we salt the roads before snowstorms. 71 Why should this be? Remember the hydration shell? When we dissolve something in water, the water molecules surround the solute particles and weakly bond to them (IMAF s). To get water to evaporate from a solution, more energy has to be put in to break the attractions between water and solute particles. So, solutions have higher and higher boiling points as we dissolve more and more solute in them. 72 Boiling Point Elevation Solute particles weaken IMF in the solvent. 1

But what about freezing? When water freezes, it has to form a crystal lattice. This is harder to do if there are solute particles in the way.

Freezing Point Depression 74 Applications salting icy roads making ice cream antifreeze cars (-64 C to 136 C) fish & insects Dissolving Electrolytes When we dissolve an

This breaking apart is called a dissociation. It is a physical change. How many moles of ions will we get if we dissolve 1 mole of CaBr 2? What about Al(NO 3 ) 2?

2 But what about freezing? When water freezes, it has to form a crystal lattice. This is harder to do if there are solute particles in the way. So, even more energy has to be removed to get the crystal lattice to form. The more solute in the solution, the lower is the freezing point. Freezing Point Depression 74 Applications salting icy roads making ice cream antifreeze cars (-64 C to 136 C) fish & insects Dissolving Electrolytes When we dissolve an electrolyte (e.g. a salt), we get more particles in solution: NaCl (s) Na + (aq) + Cl - (aq) For every mole of NaCl we dissolve, we get 2 moles of dissolved ions. This breaking apart is called a dissociation. It is a physical change. How many moles of ions will we get if we dissolve 1 mole of CaBr 2? What about Al(NO 3 ) 2? 77 Dissolving Non-Electrolytes Non-electrolytes do not dissociate in water: C 6 H 12 O 6 (s) C 6 H 12 O 6 (aq) We only get as many particles in solution as we start with when we dissolve electrolytes. highest temperature?

3 a) 100g C 6 H 12 O 6 in 500 g of water c) 100g AlCl 3 in 500g of water d) 100g MgF 2 in 500g of water d) 100g AlCl 3 in 500 g of water d) 100g AlCl 3 in 500 g of water

4 d) 100g AlCl 3 in 500g of water t = k m n t: change in temperature ( C) k: constant based on the solvent ( C kg/mol) m: molality (m) n: # of particles 86 Calculating Freezing Point of Solutions For every mole of particles dissolved in 1kg (aka 1L) of H 2 O, the freezing point of the solution decreases by 1.86 Calculating Boiling Point of Solutions For every mole of particles dissolved in 1kg (aka 1L) of H 2 O, the boiling point of the solution increases by Figure out how many moles of particles are in the solution (the molality of the solution) 2.Multiply #1 by Subtract #2 from the freezing point of H 2 O (aka 0 ) 1.Figure out how many moles of particles are in the solution (the molality of the solution) 2.Multiply #1 by Add #2 to the boiling point of H 2 O (aka 100 ) # of Particles Nonelectrolytes (covalent) remain intact when dissolved 1 particle Electrolytes (ionic) dissociate into ions when dissolved 2 or more particles What is the boiling point of a 2.5m solution of NaCl (aq)? b.p. =? t b =? k b = 0.52 C kg/mol m = 2.5m n = 2 t b = k b m n t b = (0.52 C kg/mol)(2.5m)(2) t b = 2.6 C b.p. = 100 C C b.p. = C 4

At what temperature will a solution that is composed of 0.73 moles of glucose in 225 g of phenol boil? ( The boiling point of phenol = 181.

k b = 3.60 C kg/mol m = 3.2m n = 1 t b = k b m n m = 0.73mol 0.225kg t b = (3.60 C kg/mol)(3.2m)(1) t b = 12 C b.p. = 181.8 C + 12 C b.p. = 194 C f.p. =? mol = 28g 58.

5 At what temperature will a solution that is composed of 0.73 moles of glucose in 225 g of phenol boil? ( The boiling point of phenol = C, and the boiling point elevation constant = 3.60 C kg/mol) Find the freezing point of a saturated solution of NaCl containing 28 g NaCl in 100. ml water. b.p. =? t b =? k b = 3.60 C kg/mol m = 3.2m n = 1 t b = k b m n m = 0.73mol 0.225kg t b = (3.60 C kg/mol)(3.2m)(1) t b = 12 C b.p. = C + 12 C b.p. = 194 C f.p. =? mol = 28g 58.44g/mol = 0.48mol t f =? m =0.48mol 0.100kg k f = 1.86 C kg/mol t f = (1.86 C kg/mol)(4.8m)(2) m =? n = 2 t f = k f m n t f = 18 C f.p. = 0.00 C - 18 C f.p. = -18 C Any Questions? 94 5

B. Types. Salting The Roads. A. Definition 4/21/2015. Unit 11: Solutions Lesson 11.3: Colligative Properties 68. Colligative Property

B. Types. Salting The Roads. A. Definition 4/21/2015. Unit 11: Solutions Lesson 11.3: Colligative Properties 68. Colligative Property Salting The Roads Unit 11: Solutions Lesson 11.3: Colligative Properties 68 Colligative Property A. Definition property that depends on the concentration of solute particles, not their identity B. Types