13.5 Colligative Properties

Charles Jie, Brendan Buckbee, Rajeen Amin, and Chris Swisher

Some Basics

Electrolytes dissolve in solution, Nonelecrolytes do not. Thus, electrolytes dissolved in water would ideally have twice as many particles dissolved in solution as nonelectrolytes.Raoult’s Law:   PA = vapor pressure with solute  PA° = vapor pressure without solute  XA = mole fraction of solvent in solution A PA=(XA)(PA°) Ideal Solution obeys Raoult's Law 

Boiling Point  and Freezing Point infoKb=Molal boiling-point-elevation constant DTb= change in boiling point m=molalityDTb=(Kb)(m) Kf =molal freezing-point-depression constantm=molalityDTf= change (decrease) in freezing pointDTf=(Kf)(m)   

Osmosis

Osmosis: the movement of a solvent from low solute concentration to high solute concentrationPi=(n/V)RT Pi=MRT

 Pi=osmostic pressure  M= Molarity    R= Ideal-gas Constant    T= Temperature in Kelvin

  

More info

Differences between observed and Expected changes in Temperature points  Ratio of actual value of colligative property to calculated value as a nonelectrolytei= (Tf measured)/(Tfnonelectrolyte)affected by dilution and the magnitudes of the charges on the ions

i=van’t Hoff factor ( measure of the extent of ion dissociation)

Problem 45

A. An ideal solution obeys Raoult's law (PA=XAPoA) 

B. at 60oC PoH2O= 149 torr

 assume 1 mole each of water and ethylene glycol XH2O= 1 mol H2O/(1 mol H2O + 1 mol ethylene) = 0.5PH2O=(.5)(149 torr) = 74.5 torr  However, 67 torr is the actual PH2O. Therefore, the solution is not ideal according to Raoult's law.

Problem 47

A. at 65.3 C PoH2O = 187.5 torr

 100.0g H2O x  1 mol H2O/ 18.02g H2O = 5.55 mol H2O  15.0g C12H22O11 x 1 mol/ 342.34 g = .044 mol C12H22O11PH2O= XH2O Po

H20      XH2O= 5.55 mol H2O/ (5.55 mol H2O + .044 mol)                                  XH2O= 0.992PH2O=(.992)(187.5 torr)= 186 torr                               B. at 40oC Po

H2O= 55.3 torr   500g H2O x 1 mol H2O/18.02g= 27.7 molPH2O= XH2O PoH20 (55.3 torr- 4.60 torr)=XH2O(55.3 torr)            XH2O= 0.917 = 27.7mol H2O/(27.7 mol + xmol C3H8O2)                    x= 2.51 mol C3H8O2 x 76.11g/1 mol C3H8O2    

                        x= 192 g C3H8O2 

 

Problem 49A) assume the prescence of 10g H2O and C2H5OH10.00g H2O x 1mol H2O/18.02 g H2O = 0.5549mol H2O10.00g C2H5OH x 1mol C2H5OH/46.08g C2H5OH= 0.2170mol C2H5OH 0.5549 mol+ 0.2170 mol =0.7719 mol solution    0.2170 mol     0.7719 mol      =0.2811B)Ptotal=Xa Po

a    + Xb Pob

Ptotal=(.2811) (400 torr)   +(0.7189) (175 torr) =238 torrC)XEth = (PEth)/(Ptotal) = (.2811*400torr)        =0.472                                         238 torr           

Problem 51

a) NaCl has a higher boiling point than glucose because it is a strong electrolyte and disassociates in water, which would produce twice as much dissolved particles as one mole of glucose (nonelectrolyte). Thus, NaCl would have more moles of dissolved particles and thus would have a higher boiling point.

b)  NaClDeltaT = KbmDeltaT = (2)(0.512C/m)(0.10m) = 0.10 C T(b) = 100 + 0.10 CT(b)=100.10 C C6H12O6

DeltaT = (0.512C/m)(0.10m) = 0.051 C T(b) =100+ 0.051 CT(b)=100.1C

Problem 53

ΔTF = KF · m · i ΔTF, the freezing point depression KF, the molal freezing point depression constant  m is the molality (mol solute per kg of solvent) i number of solute particles per mol

ΔTF =.040 m glycerin (C3H8O3) x 1.86 C/m x 1 = -.0744 CΔTF =.020 m KBr x1.86 C/m x 2 = -.0744 CΔTF =.030 m phenol (C6H5OH) x 1.86 C/m x 1 = -.0558 C

.030 m phenol > .040 m glycerin = .020 m KBr

Problem 57Solve for Molarity of AspirinMolarity= moles solute              liters solution

Solve for Osmotic Pressure 

   π=MRT    M= Molarity    R= Ideal-gas Constant    T= Temperature in Kelvin

 

0.64g of Adrenaline elevates the boiling point of 36.0g CCl4 by 0.49oC. What is the molar mass of adrenaline?0.49oC x  1m         x     1kg    x 36.0gCCl4 = 0.0035mol                5.02oC        1000g   

0.64g x          1          = 180g/mol              0.0035mol

Problem 59

Problem 610.150g of Lysozyme in 210mL of solution has an osmotic pressure of 0.953torr  at 25oC. What is the molar mass of Lysozyme?25oC + 273 = 298K     n = Vπ                                          RTn = .210L x 0.953torr                 = 1.1x10-5mol      62.36(L-torr/K-mol) x 298K0.150g x          1         = 1.4x104mol               1.1x10-5mol