Chapter 19: Molality and Colligative Properties HW Ch. 19 Blue Book: #1-17, 19 (on problems that are a-z, please do a and b only)

Plan of the Day Begin Ch. 19 – Notes & Examples Freezing Point Depression Lab (aka Ice Cream Lab) – 5/31 – Each student needs to bring: ½ cup milk ½ cup heavy whipping cream ¼ tsp vanilla 1 (1qt.) ziploc bag 1 (gallon) ziploc bag HW Assignment - #1-17 & 19 TEST: 6/4 Divide & conquer is the best method

Molality The volume of a solution changes with a change in temperature which alters the molarity. (ex: what happens when you boil 3 cups of water… do you still have 3 cups?) Masses, however, do not change with temperature. So… we use molality ( m) - the number of moles of solute per kilogram of solvent (mol/Kg, another concentration ratio).

Example #1 If 60.0 g of NaOH are dissolved in 1500g of water, what is the concentration of this solution using Molality? 1.Analyze the problem – what info are we given? mass of solute, 60.0 g NaOH mass of solvent, 1500 g H 2 O 2.Solve for the unknown Couple of ways to do this: 1. “Railroad-track” the whole thing all at once OR 2. Break it up… calculate number of moles of solute Convert mass of water from g  kg Plug in and solve for molality =1.00m

Example #2 Calculate the mass of ethanol, C 2 H 5 OH, which must be dissolved in g water to make a 2.00 m solution. Multiply the mass of water by the concentration (2.00 m ), then convert to grams of ethanol by using molecular mass of ethanol 750g H 2 O 2.00 mol C 2 H 5 OH 46.1g C 2 H 5 OH 1000 g H 2 O 1 mol C 2 H 5 OH Answer: 69.2g C 2 H 5 OH

Example #3 Determine the mass of H 2 SO 4 which must be dissolved in 2500 g of H 2 O to make a 4.00 m solution. Multiply the mass of H 2 O by the concentration ratio. Then convert from moles  grams of H 2 SO 4 using the formula mass of the acid. 2500g H 2 O 4.00 mol H 2 SO g H 2 SO g H 2 O 1mol H 2 SO 4 = 981 g H 2 SO 4

Example # 4 What is the percentage by mass of Mg(NO 3 ) 2 in a 2.50 m solution? The concentration involves 1000g H 2 O. Find the g Mg(NO 3 ) 2 & add these two masses together for total mass, then calculate percentage Mg(NO 3 ) mol Mg(NO 3 ) Mg(NO 3 ) g H 2 O 1 mol Mg(NO 3 ) 2 Mass of soln = g g = 1370 g soln % Mg(NO 3 ) 2 = g = 370 g Mg(NO 3 ) g H 2 O X 100 = 27.0 %

Example # 5 How many molecules of ethanol must be dissolved in g of water to make a 1.00 m solution? Multiply the mass of water by the concentration. Then multiply by Avogadro’s # /mol g H 2 O 1.00 mol C 2 H 5 OH 6.02 x molecules 1000 g H 2 O 1 mol Answer: 3.01 x molecules C 2 H 5 OH

Colligative Properties What are they? What are they? – The word Colligative means “depending on the collection” – Change the physical properties of the solvent. – Depends on the number of particles of the solute NOT which solute is used!

Colligative Properties Lowers the vapor pressure! Raises the boiling point! Lowers or depresses the freezing point! Osmotic pressure Why?

Colligative Properties When a solute is dissolved in a solvent, the vapor pressure of the solvent is reduced. The reduction depends on the number of solute particles in a given amount of solvent. The French chemist, Raoult, first discovered the vapor pressure lowering relationship experimentally in 1882 which lead to…

Colligative Properties Raoult’s Law: Any nonvolatile solute at a specific concentration lowers the vapor pressure of the solvent by an amount that is characteristic of that solvent.

Vapor Pressure Lowering The vapor pressure above a liquid is lowered due to the attractive forces of the solvent on the dissolved solute particles. Because of this, less solvent particles have the energy to transition to the gaseous state (evaporate), and therefore the vapor pressure is lower. So… The greater the number of solute particles in a solvent, the lower the VP

#1 – solvent has a large surface area to evaporate from #2 – mixed with solute = fewer solvent particles at surface #1 – solvent has a large surface area to evaporate from #2 – mixed with solute = fewer solvent particles at surface Beaker #1Beaker #2 Pure SolventSolution Which one has lower VP?

Boiling Point Elevation Similar factors (as with the vapor pressure lowering), contribute to the increase of the boiling point of a solvent. – The more solute particles the higher the BP (the lower the VP) Practical application – adding salt to water to increase the BP of water to cook foods.

Boiling Point Elevation

Freezing Point Depression Freezing occurs when the particles no longer have the energy to overcome their interparticle attractive forces – they organize and solidify (molecules slow way down, loss of kinetic energy). Adding solute to a pure solvent lowers the FP! – WHY? Because the solute interferes with the solvents interparticle attractions, therefore the solid forms at cooler or lower temperature. So… the FP of a solution is always lower than the FP of a pure solvent.

Freezing Point Depression

0oC0oC 100 o C ___ = Pure Solvent ---- = Solution

Osmotic Pressure What is osmosis? The amount of additional pressure caused by the water molecules that move into a concentrated solution is called osmotic pressure. (The diffusion of water) This pressure depends on the number of solute particles in a given volume of solution.

As water is moving  the pressure exerted by the additional water molecules, osmotic pressure, is increasing on the left side of the semipermeable membrane. Higher osmotic pressure on left, lower osmotic pressure on right.

Colligative Properties (now the math) The change in the freezing and boiling pts varies directly with the concentration of particles. Molal freezing pt constant: 1.86C˚ for water. Each mole of solute causes the freezing pt of water to drop by this much. Molal boiling pt constant: 0.512C˚ for water. Each mole of solute causes the boiling point to rise by this much.

Colligative Properties These can be used to determine: The freezing point of the water The boiling point of the water The molecular mass of the solute from the freezing point or the boiling point (see table 19-1 for other constants)

Colligative Properties Ex. 3 Calculate the freezing point of a solution containing 5.70 g of sugar, C 12 H 22 O 11, in 50.0 g of water. (Molal freezing pt constant: 1.86C˚ for water. ) Convert grams of solute per gram of water to moles of solute per kg of water (molality). Then multiply by the conversion ratio to obtain the change in FP 5.70 g C 12 H 22 O g H 2 O 1 mol C 12 H 22 O C˚ 50.0 g H 2 O 1kg H 2 O 342 g C 12 H 22 O 11 1 m = 0.620C˚, To determine the FP, subtract this from the FP of water 0 o C – = o C

Calculating Molecular Mass Ex. 4 When 72.0 g of dextrose were dissolved in g of water, the boiling point of the solution was observed to be ˚ C. What is the molecular mass of dextrose? Step 1: determine the molality of the solution 100 o C ˚C = 2.05˚C determine the  T b 2.05 o C m = 4.00 m ˚C molal boiling pt. constant for H 2 O Step 2: determine the grams per mole 72.0 g dextrose 1 kg H 2 O = 180 g kg H 2 O 4.00 mol mol

One last thing: Colloids Colloids are not true solutions, but special types of mixtures that behave like solutions. – There are two parts, the dispersed phase and continuous phase. Dispersed phase has particles from 1 to 100 nm in size and remain dispersed by the random motion of the molecules (kinetic energy). Any particle larger than 100 nm will usually settle out over time.

Mark Rosengarten videos Antifreeze, Electrolytes…: – wi2E&feature=related wi2E&feature=related BP elevation & FP depression – so&feature=related so&feature=related Molality – BDXU&feature=related BDXU&feature=related

Freezing Point Depression Lab Do the lab… eat the ice cream Do the lab… eat the ice cream – Complete the problems on the back of lab sheet – Complete the problems on the back of lab sheet Hand out Ch. 19 Test Review Hand out Ch. 19 Test Review HW time HW time Ch. 19 Test next Class!!! Ch. 19 Test next Class!!!

Plan for the Day ~30 min Review ~30 min Review Turn in NB’s Turn in NB’s Chapter 19 Test Chapter 19 Test

Test Addition On test… On test… #2 c. rate of diffusion (osmotic pressure) #2 c. rate of diffusion (osmotic pressure)