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Chapter 14

Colligative Properties some properties of solutions depend on the number of solute particles per solvent molecules
Units of Concentration for Colligative Properties molality (m)=moles of solute/kilogram of solvent
What is the molality of a solution that contains 5.67g of glucose (c6h12o6) in 25.2g of water? Change 5.67g c6h1206 to moles Change 25.2g of water to kg Molality M=moles solvent/kg solvent
Mole fraction (Xa)=moles of substance A/ total moles of solution
What is the mole fraction of glucose in a solution that contains 5.67 of c6h12o6 in 25.2g of water? first- change every quantity to moles. then add them together second-mole fraction
Weight % and mass % mass % of solute= mass of solute/mass of solution *100
if a solution is labeled 2.40% sodium acetate (nac2h3o2) and there is 425g of solution. how many grams are na and acetate and how many grams are water? 1-mass % of solute- plug in and solve then grams of solution=grams of solute + grams of solvent---solve
solubility the amount that will dissolve in a given quantity of water at a given tmeperature
NaCl is soluble in water. how much NaCl can dissovle in a given amount of water? [solubility-36.0g/100g water at 20.c] if we dissolve 36.0g of NaCl in 100g of water at 20.c, we wold have a saturated solution
If we dissolve 30g of NaCl of water at 20.c, our solution would be unsaturated unsaturated solution-holding less solute than it is capable of holding at a given temp
supersaturated solution it is holding more solute than the maximum at a given temp
miscible 2 liquids are mixed to form a solution
immiscible 2 liquids that do not mix
dissolving process exothermic
to be soluble, an ionic compound will have have an enthalpy of solution that is exothermic or only slightly endothermic
Lattice Energy and Hydration Energy ion charges/ion size -greater charge, stronger l.e. and h.e. -smaller ions, stronger l.e. and h.e.
heat enthalpy of solution can be calculated by taking the difference between the enthalpy of formation of the compound (<|Hf.) and the enthalpy change for dissolving process (<|H. solution)
Determine the enthalpy of solution for NaCl. the <|Hf. for NACl is -411.1 kj/mol. The <|H. solution for NaCl is +38 kj/mol <|hf.-<|H. soln=net dissolving process to form NaCl -411.1kj- +38 kj = -414.3 kj (1m)
Enthalpy of Solution when ionic cmpd dissolves in water, 2 steps (enthalpy is sum): 1- ions in ionic lattice must separate. energy must be added for this to happen (endo) the force of attraction bt ions must be overcome
Enthalpy of Solution when ionic cmpd dissolves in water, 2 steps(enthalpy is sum): 2-hydration must occur. ions become surrounded by water molecules. must be a strong ion-dipole attraction (exo)
KF+H2O-->K+ + F- 821 KJ are added to separate 1 mole of KF into ions. 837 kj of energy is released when the enthalpy ions are hydrated. what is the enthalpy of soslution <|H soln=<| lattice + <|hydration -16kj=821kj+ - 837 kj ^break lattice^given off
Factors that affect solubility pressure&temp both affect the solubility of gases in liquids temp only affects the solubility of solids in liquids
pressure effects: Henry's law -the solubility of a gas in a liquid is directly portional to the gas pressure (^p,^s) (\/p,\/s)
Henry's law Sg=KhPg s=solubility of gas (mol/kg) pr (mol/L) Kh=henry's constant (varies w each gas) Pg= pressure on the gas (units must agree w henry's constant) (find examples)
temp effects: solubility of all gases in water decrease w/ increasing temp ^T\/s (gas) solubility of solids in water is also affected by temp. most solids will become more soluble as temp. increases
Le Chatelier Principle/Co2(g)-->Co2(s) when a system is in equilibrium and is disturbed by changes in temp, pressure, or concentrations. the system will shift to counteract the change -*adding pressure- more co2 to dissolve. lowering pressure-more co2 to go back to being (g)
colligative properties depends on # of solute particles in solution
first colligative property-vapor pressure of a solution -v.p. of volatile solvent will be lowered by adding a non volatile solute ([] of solute^,Vpsoln\/) -vp lowering of solvent is c.p. = to vp of pure solvent -the vp of the solution
Raoult's Law states the vp of solvent over solution = to vp of pure solvent (P.solvent) times the mole fraction of solvent (Xsolvent) ex. Pwater= (Xwater)(P.water)
2nd Colligative property-boiling point elevation normal boiling point of liquid is at temo at which its vp equals=atm pressure -the addition of nonvolatile solute will reduce the vp, so the temp must be increased beyond the normal b.p. to achieve a vp=atm pressure (1atm)
to calculate for boiling point elevation: <|Tb=1Cb*M 1Cb=boiling constant of solvent M=molality of solution *1Cb for water is .51. c/m **once you solve for the bp of elevation you must add this to the solvents normal bp to get solutions
ex. What is the BP of a solution that contains 30g of area ((NH2)2CO) in .500kg of water 1st-solve for the molality of a solution next solve for the <|Tb (BP elevation) Now add the bp elevation to the normal bp
ex. How many grams of ethylene glycol must be added to 125 g of water to raise its bp 1.5.C? 1st-solve for molality of soln 2nd-solve for moles of ethylene gylcol using molality equation 3rd-change moles to grams
3rd colligative property-freezing pt depression when solute is dissolved in a solvent the FP of the soln will be lower than that of the pure solvent
to calculate the fp depression (<|f) use this formulka: <|Tf=Kf*M -once you solve for the FP depression, you must subtract this from the solvent's normal FP to get the solution's new FP
FP depression problem worked like BP elevation problems when you calculate <|Tb you just add this to the solvents normal BP and when you calculate <|Tf you must subtract this from the solvent's normal FP
4th colligative property-osmotic pressure -osmosis-the movement of solvent molecules through a semipermeable membrane from a region of lower solute concentration to a region of higher solute concentration
4th colligative property-osmotic pressure -osmotic pressure-the pressure that is applied to a soln that just stops osmosis. osmotic pressure (π) can be calculated using this formula: π= MRT
π=MRT M=molarity of the solution r=gas constant (.821 atm*l/mol*k) t=temp in kelvin
ex. Calculate the osmotic pressure of a soln that contains 5.0g of sucrose in 100mL of soln at 20*c 1st- solve for the molality of soln now use osm. pressure formula -even soln w low solute concentrations have significant osmotic pressure
using osmotic pressure calculation to solve for the mm of an unknown soln: ex. what is the mm of a solute when 7.68mg is add to make 10ml of soln. the osmotic pressure is 26.57mmHg at 25.c 1st change mg to g and then solve for moles now solve for moles using molarity formula now solve for mm
colligative properties of solutions that contain ions -only depend on the # of dissolved particles in soln
If the solute is an ionic compound determine the # of ions in the formula
Colloids -dispersion of particles throughout another substance or soln -are not the soln -ex. jello, milk, fog, aerosis, smoke, whipped cream -contain larger particles than settle out -can scatter light (tyndall effect) -can be hydrophobic ("water fearing")
Hydrophobic colloids -generally unstable -include oils, fats, soil
Hydrophillic -proteins, milk, starch
Substances such as soaps that affect the properties of surfaces and affects the interaction between 2 phases are called surfactants -detergents help to break the surface tension of water and interact with polar substances and non-polar substances
Created by: dixxfranks



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