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Ch 14 Solutions

by: Alexi Martin

Ch 14 Solutions CHEM 1200

Alexi Martin
GPA 3.58

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This is the last chapter that will be covered on exam 1, it covers ch 14 from dr ma's lecture
Chemistry II
Dr. Alexander Ma
Class Notes
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This 4 page Class Notes was uploaded by Alexi Martin on Monday February 22, 2016. The Class Notes belongs to CHEM 1200 at Rensselaer Polytechnic Institute taught by Dr. Alexander Ma in Spring 2016. Since its upload, it has received 18 views. For similar materials see Chemistry II in Chemistry at Rensselaer Polytechnic Institute.


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Date Created: 02/22/16
Ch 14 Solutions and Colligative Properties    Solutions  ­homogeneous mixture, can contain multiple materials  ­▯S mix> 0, mixing will increase entropy  ­cannot see parts­ ​omogenous ​ (called solutions)  ­component of solution changes state­s​olute  ­component keeps state ​solvent­ major portion  ● solutions contain kg amalgams, metals= alloys  ­ brass= Zn+Cu  Colligative Properties  ­ value depends on # of solute particles and what they are and/ or their concentration  ­ difference in value of solution and substance different attractive forces and solute  occupying solvent positions  Vapor Pressure Solution  ­solvent above solution has a lower vapor pressure than of a pure solvent  ­solute replaces solvent molecule at the surface of the solution  ­add nonvolatile solute, decreases the opportunity to vaporize and decreases the vapor  pressure  Thirsty Solutions  ­ example: salt water, the concentration of the solution draws solvent towards it, draws  pure solvent vapor and mixes, lowers vapor pressure  Raoult's Law  ­ Vapor pressure of volatile solvent above a solution:  Psolvent=XsolventP^о  ­ mol x<1 Vapor pressure will always be less than vapor pressure in solvent  ­ mole fraction X=mol of A/mol of A+B  example 1:​  99.5 C12H22O11 300.0 mL of H2O   99.5g/342.30g= 0.2907 mol of C12H22O11 300ml=300g/18.02g=16.65 mol of H2O  mol fraction of H2O= 16.65/0.2907+16.65= 0.1387 mol fraction H2O P^o= 23.8 torr  (using equation from above) 0.9828(23.8)= ​PH2O=23.4 torr  example 2: ​ uses same equation as above example   25 g/180.2g=0.1387 mol 215g/18.02g= 11.93 mol H2O  P^o=92.5 torr  mol fraction of H2O=11.93/0.1387+11.93= 0.9885  0.9885(92.5)=​91.4 torr=PH2O  *example 1 & 2 are for nonvolatile solutions*  Vapor Pressure Lowering   ­for volatile solutions  ­vapor pressure solvent in solution lower vapor pressure of pure solvent  ­vapor pressure solution is proportional to the amount of solvent in solution  ­vapor pressure lowering:   △P=P^osolvent­P^osolution=Xsolute(P^osolvent)  1  Raoult's Law Volatile Solute  ­both can evaporate, both will be vapor solute & solvent  ­ vapor pressure of solute and solvent is given by:  Ptotal=Psolute+Psolvent ​­> uses equation above (Raoult’s Law) for both solute and solvent  ­solvent vapor pressure lower than solute vapor pressure  example 3: ​ .95g/76.15g=0.05187 mol     2.43/58g=0.04184 mol  0.05187/(0.04184+0.05187)=0.5835 to get other mol fraction 1­0.5535  PCS2=(0.5535)515= 285 torr 0.04665(332)=148 torr adding them together yields 148+285=​ 443  torr  example 4:  0.500/0.500+0.250(440)=293.3 torr  0.250/0.500+0.25(44.6)= 14.87 torr  293.3+14.87= ​308 torr  ~clicker problem~ ​  2 solutions 10 g of glucose 1 L H2O, 10 g sucrose 1 L of water, what is  true? B sucrose has higher vapor pressure, a ​ higher molar mass yields a higher vapor  pressure  Ideal vs. Nonideal solutions  ­ according to Raoult's Law­ ideally solvent­solute interactions =  solute­solute+solvent­solvent  ­ solute dilutes solvent  ­ solute and solvent are stronger/weaker than broken interactions between molecules is  non ideal  Other colligative properties related to lower vapor pressure  ­ vapor pressure lowers at all T  ­ boiling point­ requires more energy & freezing point lower temperature to freeze  Freezing NaCl(aq)  ­ water freezes at 0 degrees Celsius, equilibrium  ­ add salt/disrupt equilibrium, solute changes freezing point, freezes more slowly until it  returns to equilibrium at a lower temperature  Freezing Point Depression  ­ Freezing point solution is lower than the freezing point of the pure solvent, melting point  solution is also lower  ­ ΔT freezing >0  ­ freezing point depression constant Kf: depends on solvent   ­ molality=mol solute/kg solvent : does not vary with T because its based on masses  FPsolvent­FPsolution=ΔTf=m*Kf  example 5: ​ .7 (1.86) = 3.2 degrees celsius 0­Fsolution=3.2  Fpsolution= ­3.2 degrees celsius  example 6: ΔTf=mKf 1.94=m(1.86) m=1.043=x/0.080 x=0.0344 mol MM=g/mol  12/0.08344=​144g  example 7: ​5.5­4.1=1.3=5.12  m=0.273=x/0.1 x=0.027 mol 5/0.027 =1 ​ 83 g/mol   Boiling point Elevation  ­ boiling point rising solution than pure solvent is nonvolatile, Tb>0  BPsolution­BPsolvent=ΔTb=mKb  2  example 8:  105­100=5=m(0.512) m=9.97=9.922 mol (62.07) =​ 6.1X10^2  example 9: 1/92.09 0.011/0.054 0.201(0.512)=0.1   100.103 degrees celsius  Osmosis  ­ flow of solvent from solution of a lower concentration to a solution of a higher  concentration  ­ semipermeable membrane allows flow of solvent not solute  Osmotic Pressure  ­ amount of pressure needed to keep osmotic flow active  ▯=MRT ​ where pi is in atm, M is molarity= mol of solute/L solution R =0.08206  mg= 10^­3 g  example 10:  2.45/760=298(x/0.01) 1.318x10^­4=1.318x10^­6  5.87x106­3/1.1318x10^­6 =​ 4.45x10^3 atm  example 11: 1.32x10^­3=298(0.08206)M 5.398x10^­5=x/.1    0.0750/5.398x10^­6=1 ​.389x10^4  g/mol  Van’t Hoff Factors  ­ Ionic compounds can make multiple particles  ­ add i to all colligative property equations  ­ i ratio mol solute: mol formula units dissolved  ­ i< theoretical due to ion guessing  ­ causes ΔT to be lower than expected  ~ clicker problem~ ​largest osmotic pressure?0.5 M FeCl3​, 4 ions will dissolve  example 12: van hoff factor? use freezing point depression equation  i=0.27/0.050(1.86)=2.9=i  example 13: P solution? iXH2OP(degrees) H2O  0.927/3(0.927+0.102) 0.7518(1.18) ​ solution=88.8 torr  ~clicker problem~​ higher BP higher # of particle​.05m NaCl  example 14: ​ se boiling point elevation equation  0.512, 0.1711/0.0546=m 2(3.169)(0.512) BP=100 degrees celcius so B​ P=103.245 degrees  celsius  Solubility   ­solute­ solvent=soluble  ­does not dissolve insoluble  ­solubility depends on tendency of mixing and intermolecular forces  Spontaneous mix­ solute and solvent  Mixing and solution­entropy  ­ gas and gas mixed will spontaneously no intermolecular forces from lower energy to  higher energy  Mixing & the Solution  ­entropy measures energy dispersion throughout the system, energy spreads out over a certain  volume, the gas expands, an increase in entropy  Intermolecular Forces and the solution process  3  ­ energy changes in the formation of most solutions also involves differences in attractive  forces  ­ solvent and solute to mix, must overcome: all solute­solute attractive forces; some  solute­solvent attractive forces (endothermic)  ­ energy to do this new solute­solvent attractions will be exothermic  Relative interactions  ­solute to solvent > solute­solute+ solvent­solvent: exothermic ΔH<0 ΔS>0 solution forms  ­solute to solvent= solute­solute +solvent­solvent: ΔH=0 ΔS> 0 solution forms  ­solute to solvent < solute­solute +solvent­solvent ΔH>0 ΔS<0 solution may not form  (endothermic) temperature dependant  ­solute to solvent weaker + solute+solvent will form if the energy­energy barrier is small enough  to overcome  ­ ΔG­ΔH­TΔS  Solubility  ­ gases will always mix, 2 liquids soluble if they are miscible (alcohol and water),  immiscible (oil and water)  ­ max solute can be dissolved in a solvent­ solubility varies with pressure and temperature  Will it Dissolve?  ­ like dissolves like, if they have a similar structure  ­ polar ion in polar, nonpolar in nonpolar  ­ Vitamin C in water, vitamin K3 will not in water  Heat of Solution  ­ NaOH dissolves heat is released: exothermic  ­ NH4NO3 dissolves in water endothermic  ­ energetics of solution formation is the enthalpy of the solution  To make a solution  1. overcome all attractions with solute ΔH solute endothermic  2. overcome some attractions between solvent ΔHsolvent endothermic  3. Form new attractions solute­solvent ΔHmix exothermic  ΔHsolution=ΔHsolute+ΔHsolvent+ΔHmix      4 


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