Ch 14 Solutions
Ch 14 Solutions CHEM 1200
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Popular in Chemistry
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 statesolute 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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