Drill (pd 3) 5/11/2015 What are the 3 rules for determining solvation (what dissolves what)? Determine the number of grams of solute needed to make a 0.182 m solution of sucrose (C12H22O11) in 275 g of water.

Answers Like dissolves like: Polar solutes dissolve in polar solvents Nonpolar solutes dissolve in nonpolar solvents Ionic compounds are soluble in polar solvents molality = amount of solute (mol) 0.182 m = X moles mass of solvent (kg) 0.275 kg X = 0.05 moles x 342.3 g sucrose = 17.1 g sucrose I mole sucrose

Announcement HW – Solubility Graph Test – Thursday or Friday this week Lab – Tomorrow or Wednesday this week – wear closed toe shoes. Lab Notebooks – project grade

Drill 5/11/2015 Why do you think salt lowers the freezing point of ice on the road?

Answer Addition of solute to form a solution stabilizes the solvent in the liquid phase, and lowers the solvent chemical potential so that solvent molecules have less tendency to move to the gas or solid phases. Liquid solutions slightly below the solvent freezing point become stable meaning that the freezing point decreases.

Agenda Pass forward Molality WS (all periods), Making a Solubility Curve Graph and WS Notes on Polarity (WS) Notes on Colligative Properties (WS)

SWBAT Calculate the freezing point depression and boiling point elevation of a solute both in the classroom and the lab.

Homework DUE Polar vs Nonpolar WS

Announcement Test – Thursday or Friday next week Lab – Tuesday or Wednesday next week – wear closed toe shoes.

Colligative Properties

What is a colligative property? A property that depends on the concentration* of solute particles – NOT the identity of the particle. Vapor Point Reduction Boiling Point Elevation Freezing Point Depression Osmotic Pressure * Concentration will be in terms of molality (m)

Vapor pressure - is pressure caused by molecules in the gas phase that are in equilibrium with the liquid phase. (in a closed system particles go back and forth between phases)

Vapor Pressure Reduction The pressure of the vapor over a solvent is reduced when a solute is dissolved in it. Vapor pressure reduction is directly related to concentration of the solution.

Why does this occur? Increasing solute particles reduces the proportion of solute to solvent. Fewer solvent particles will be available to leave the solution and enter the gas phase. The solution will remain a liquid over a larger temperature range.

Phase Diagram Shows us what phase the solvent will be in at a given temperature and pressure. If we add a solute that lowers the vapor pressure, then our phase diagram adjusts…

Boiling Point Elevation Boiling point – the temp. at which vapor pressure of a liquid is equal to atmospheric pressure. The boiling point of a solvent is raised when a solute is dissolved in it. It is directly related to the concentration of the solute.

Tb = Kb  m  # of particles The change in boiling point is proportional to the molality of the solution: Tb = Kb  m  # of particles Kb is the molal boiling point elevation constant, a property of the solvent. It has the units ºC/m. Tb is the boiling point elevation. It is the difference between the boiling points of the pure solvent and that of the solution. # of particles: nonelectrolytes = 1 electrolytes = # of ions produced in solutions

Freezing Point Depression Freezing point – the temp. where vapor pressure of the solid and liquid phases are the same. The freezing point of a solvent is lowered when a solute is dissolved in it. Freezing point depression is directly related to concentration of a solution.

Tf = Kf  m  # of particles The change in freezing point can be found similarly: Tf = Kf  m  # of particles Here Kf is the molal freezing point depression constant of the solvent. Tf is the freezing point depression. It is the difference between the freezing points of the pure solvent and that of the solution. # of particles: nonelectrolytes = 1 electrolytes = # of ions produced in solutions

Practice Problem NaCl is used to prevent icy roads and to freeze ice cream. What are the boiling point and freezing point of a 0.029 m aqueous solution of NaCl? Kb = 0.512 C/m Calculate Tb based on number of particles in the solution. Tb = Kb  m  # of particles Tb = (0.512) (0.029) (2) = 0.030 C Boiling Point = 100 C + 0.030 C = 100.030 C

Practice Problem NaCl is used to prevent icy roads and to freeze ice cream. What are the boiling point and freezing point of a 0.029 m aqueous solution of NaCl? Kf = 1.86 C/m Calculate Tf based on number of particles in the solution. Tf = Kf  m  # of particles Tf = (1.86) (0.029) (2) = 0.11 C Freezing Point = 0.00 C - 0.11 C = -0.11 C

Drill #5 4/22 & 23/14 Using prefix meanings as a clue, which direction does water move in hypotonic, hypertonic and isotonic solutions? Hint: the prefix refers to concentration of solute outside cell

SWBAT Describe the properties of hypertonic, hypotonic and isotonic solutions. Calculate the freezing point depression and boiling point elevation of a solute both in the classroom and the lab.

Agenda Finish Notes on Colligative Properties - Osmosis Antifreeze Lab Calculations

Homework DUE Effect of a Solute on Freezing and Boiling Points WS

Homework Colligative Properties Quiz – next class! 15-4 Review and Reinforcement - #1 - #11 15-4 Practice Problems - #1, 3, 4, 7, 9, 11,12,15,16, and 18.

Osmosis Water diffuses from an area of high concentration to an area of lower concentration

Osmosis 3 types of osmosis (based on amount of solute outside the cell) Hypertonic solution (higher) Hypotonic solution (lower) Isotonic solution (equal)

Hypertonic Solution Higher water concentration inside of the cell compared to the outside More solutes on the outside of cell Water diffuses out of cell Cell shrinks

Hypotonic Solution Higher water concentration on the outside of cell than inside Less solutes on the outside of cell Water diffuses into cell Cell gets bigger

Isotonic Solution Same water concentration outside and inside of the cell Same solute concentration on outside and inside of cell Water moves in and out at the same time Cell stays the same size

Pre-Lab Questions We will begin with pre-lab question 5 5. Why is calculation step 5 necessary? Hint: What are you trying to calculate?

6. Write the mathematical expression for molality.

7. Write the mathematical expression that relates the math, number of moles and molar mass of a substance Hint: What is the formula for molar mass?

Hint – Combine the equations from step 6 and 7 to determine the equation. 8. Using these two expressions, derive an equation for calculating the molar mass of a sample if you know its mass and the freezing point depression it causes for water.

Post Lab Calculations 1-2 Record the freezing point depression of your solutions 3-4 Determine the molality (m) of each solution Hint: What equations do you know for colligative properties

Post Lab Calculations Cont. To find the molar mass of antifreeze, you first need to calculate the number of grams of antifreeze per 1000 grams of solvent for the solutions Hint: What concentrations were solutions 1 and 2? What is the ratio between the solute and the solvent?

Post Lab Calculations Cont. Find the molar mass of antifreeze Use the equation you derived in the pre-lab

Complete the Critical Thinking Questions Questions not completed in class will become your homework

Assignment Effect of a Solute on Freezing & Boiling Points Worksheet #1-4 Pre-lab Questions

Chapter 14 Test: this Friday, April 27 Topics include: Types of Mixtures – notes Solution Concentrations – molarity, molality and dilution equations – problems Factors affecting solvation - notes Colligative Properties – boiling & freezing pts - notes Osmosis – hypertonic, hypotonic, & isotonic solutions - notes