Chapter 14
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| Colligative Properties | some properties of solutions depend on the number of solute particles per solvent molecules
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| Units of Concentration for Colligative Properties | molality (m)=moles of solute/kilogram of solvent
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| 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
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| Mole fraction | (Xa)=moles of substance A/ total moles of solution
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| 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
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| Weight % and mass % | mass % of solute= mass of solute/mass of solution *100
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| 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
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| solubility | the amount that will dissolve in a given quantity of water at a given tmeperature
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| 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
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| 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
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| supersaturated solution | it is holding more solute than the maximum at a given temp
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| miscible | 2 liquids are mixed to form a solution
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| immiscible | 2 liquids that do not mix
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| dissolving process | exothermic
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| to be soluble, | an ionic compound will have have an enthalpy of solution that is exothermic or only slightly endothermic
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| 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.
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| 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)
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| 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)
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| 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
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| 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)
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| 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
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| Factors that affect solubility | pressure&temp both affect the solubility of gases in liquids
temp only affects the solubility of solids in liquids
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| pressure effects: Henry's law | -the solubility of a gas in a liquid is directly portional to the gas pressure
(^p,^s) (\/p,\/s)
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| 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)
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| 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
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| 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)
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| colligative properties | depends on # of solute particles in solution
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| 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
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| 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)
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| 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)
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| 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
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| 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
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| 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
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| 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
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| 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
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| 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
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| 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
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| 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
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| π=MRT | M=molarity of the solution
r=gas constant (.821 atm*l/mol*k)
t=temp in kelvin
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| 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
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| 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
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| colligative properties of solutions that contain ions | -only depend on the # of dissolved particles in soln
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| If the solute is an ionic compound | determine the # of ions in the formula
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| 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")
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| Hydrophobic colloids | -generally unstable
-include oils, fats, soil
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| Hydrophillic | -proteins, milk, starch
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| 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
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