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Study Guide – Properties of Solutions and Colligative Properties

Concentration units are usually in terms of Molarity. However, sometimes units of Normality, molality, Mole fraction or mass percent are used. These formulas are

Molarity (M) = __moles of solute

liter of solution

molality (m) = _moles of solute

kilogram of solvent

mass percent = grams solute X 100

grams of solution

Normality (N) = equivalents of solute

Liter of solution

Mole fraction (X) = number of moles of component

Total number of moles

Example: A solution is prepared by adding 5.84 grams of formaldehyde, H2CO, to 100.0 grams of water. The final volume of the solution was 104.0 mL. Calculate the molarity, molality, mass percent and mole fraction of the formaldehyde in the solution.

Molarity =

Molality =

Mass percent

Mole fraction

Normality

Solubility

Likes dissolve Likes - describe what is meant by this statement

Henry’s Law

Pressure has little effect on liquids and solids. However, pressure does effect gas solubilities independent of the gas structure.

Henry’s law P = kC applies to gases at low concentrations that do not react with the solution

P = partial pressure of gaseous solute above the solution in atm

k = a constant for a particular solution

C = concentration of the dissolved gas

Example: The solubility of oxygen is 2.2 X 10 -4 M at 0o C and 0.10 atm. Calculate the solubility of oxygen at 0o C and 0.35 atm.

Temp is constant so P1 = k = P2

C1 C2

Calc C2

Vapor Pressure of solutions

The behavior of nonvolatile solutes is described by Raoult’s law

P soln = X solvent Po solvent

Where P soln = vapor pressure of solumtion

X solvent = mole fraction of solvent

Po solvent = vapor pressure of the pure solvent

Example: Raoult’s Law with nonelectrolyte solute

Glycerin, C3H8O3 is a nonvolatile liquid. What is the vapor pressure of a solution made by adding 164 grams of glycerin to 338 mL of H2 O at 39.8oC is 54.75 mmHg and its density is 0.992 g/mL

Solve for moles of glycerin =

Solve for moles of water =

Solve for mole fraction =

Solve for Pressure of solution using : P soln = X water Po water

Example: Raoults’s Law with electrolytes

What is the vapor pressure of a solution made by adding 52,9 grams of CuCl2 , a strong electrolyte, to 800.0 mL of water at 52.0 oC is 102.1 torr and its density is 0.987 g/mL

CuCl2 (s) → Cu+2 (aq) + 2Cl- (aq)

Each mole of CuCl2 dissociates to form 3 moles of solute. To determine the new vapor pressure, we must calculate the mole fraction of water, then multiply by the vapor pressure of pure water.

Solve for moles of CuCl2

Solve for total moles of solute

Solve for moles of water

Solve for mole fraction

Solve using Raoult’s law: P solution = Xwater Po water

Example: Finding Molar Mass with Raoult’s Law

At 29.6 oC pure water has a vapor pressure of 31.1 torr. A solution is prepared by adding 86.7 gram of “Y” , a nonvolatile nonelectrolyte to 350.0 grams of water. The vapor pressure of the resulting solution is 28.6 torr. Calculate the molar mass of Y.

(manipulate the equation P = XPo to solve for X = P to determine Xwater )

Po

Solve for mole fraction of water using above suggestion

Solve for moles of water from grams in problem

Use the mole fraction equation and sub in values for mole fraction and water moles to find moles of Y

Molar Mass of Y = given grams Y

moles of Y

Colligative Properties: Boiling point Elevation and Freezing Point Depression, Osmotic Pressure

Colligative properties depend only on the concentration of the solute particles not the nature of the particles. Electrolytes dissociate into more than one particle, Nonelectrolytes do not dissociate.

Van’t Hoff Factor (i)

In a dilute solute concentrations, the number of particles dissociated multiplies the effect on the colligative properties. The high the concentration, the less likely it is for this statement to hold true.

Solute van’t Hoff factor

NaCl 2

C6H12O6 1

Boiling point Elevation

ΔT = Kb msolute

ΔT is the boiling point elevation in Celcius

Kb is molal boiling point elevation constant of the solvent ( in oC kg/mol) Find constants in textbook

M is the molality of the solute (in mol solute/kg solvent)

Example: a solution is prepared by adding 31.65 g of sodium chloride to 220.0 mL of water at 34.0oC ( density = 0.994 g/mL) Kb for water is 0.51oC kg/mol) . Calculate the boiling point of the solution, assuming complete dissociation of NaCl in solution (which is not totally true)

Calculate molality of NaCl

Since NaCl dissociates into 2 particles it has a van’t Hoff factor (i) of 2. With electrolytes, the formula is ΔT = iKb msolute.

Calculate ΔT

Add the ΔT to the original boiling point

Freezing point depression

Follows the same format with two differences. The Kf is the molal freezing point constant of the solvent. After calculating the Freezing point depression subtract the ΔT from the normal freezing point

ΔT = iKf msolute

Example: How many grams of glycerin ( C3H8O3) must be added to 350.0 g in order to lower the freezing point to -3.84oC (Kf for water = 1.86 o kg/mol) ?

Osmotic Pressure

Osmosis is the flow of solvent into a solution through a semi permeable membrane. Osmotic pressure is the pressure that just stops the osmosis. It is a colligative property dependent on the concentration of the solute, not its nature.

The relationship is given by: π = MRT

π is the osmotic pressure ( in atm)

M is the molarity of the solute ( in mol/L)

R is the universal gas constant ( 0.08206 atm L/K mol)

T is the temperature(in K)

Example – Osmotic pressure

The osmotic pressure of a solution containing 26.5 mg of aspartame per liter is 1.70 torr at 30o C. Calculate the molar mass of aspartame.

Change the pressure from torr to atm π =

R = 0.08206 atm l/mol K

T = 30 + 273 = 303 K

Rearrange the formula π = MRT to solve for Molarity

Solve for molar mass using the Molarity M = Mol = grams/ Molar Mass

Liter Liter

Terms to know: osmosis, hypertonic, hypotonic, isotonic, reverse osmosis, crenation, hemolysis, miscible, immiscible, dimmer, electrolyte, nonelectrolyte, solvent, solute, solution, colloid, Tyndall effect

Converting molality to Molarity

Example:

Calculate the Molarity of a 0.396 molal glucose solution. The molar mass of glucose is 180.2 g/mol and the density of this solution is 1.16 g/mL

• Use molality to convert to grams of solute

• Add 1000 g of H2O

• Calculate volume of the solution using the density

• Calculate the Molarity

Example: Change Molarity to molality

A 2.45 M solution of CH3OH has a density = 0.976 g/mL.

What is the molality of the solution?

• Using 1 liter of solution and the density, find the grams of solution

• Find the grams of solute from the Molarity

• Subtract the grams of solute from the grams of solution to find the grams of solvent

• Calculate the molality

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