Calculations Involving Colligative Properties

1. Calculations Involving Colligative Properties Freezing Point Depression and Boiling Point Elevation Calculations

2. Objectives • When you complete this presentation, you will be able to • calculate the freezing point depression or boiling point elevation of a solution when you know its molality • calculate the molality of a solution when you know its freezing point depression or boiling point elevation

3. Introduction • We now understand colligative properties. • To use this knowledge, we need to be able to predict these colligative properties. • Freezing Point Depression • Boiling Point Elevation

4. Introduction • We also need to use a different kind of concentration determination. • Instead of molarity, we will use • molality, m • mole fraction, X

5. Molality • Molarity - we measure the number of mols of solutein the volume of the solution. • Msolution = nsolute/Vsolution • Molality - we measure the number of mols of solutein the mass of solvent. • msolution = nsolute/msolvent

6. Molality • msolution = nsolute/msolvent • The mass of the solvent is measured in kilograms, kg. • 1 mole of solutein 1,000 g of solventgives a 1 msolution.

7. Molality Example 1: Find the molality of 87.66 g of NaCldissolved in 2.500 kg of H2O. mNaCl = 87.66 g mH2O = 2.500 kg MNaCl = 58.45 g/mol nNaCl = mNaCl/MNaCl m = nNaCl/mH2O First, we write down the known values. To calculate molality, m, we need to know the number of mols of NaCl. We only know the mass of NaCl. Therefore, we calculate the number of mols of NaCl. Now, we can calculate the molality, m of the solution. This value comes from the sum of the atomic masses of Na, 22.99 g/mol, and Cl, 35.45 g/mol. We substitute in known values … … and do the calculation. We put in the known values … … and do the calculation. = (87.66 g)/(58.44 g/mol) = 1.500 mol = (1.500 mol)/(2.500 kg) = 0.600 mol/kg

8. Molality Example Problems: • Find the molality of 100.0 g of NaOH in 1.500 kg of H2O. • Find the molality of 32.00 g of KClin 0.5000 kg of H2O. • Find the molality of 142.5 g of CuCl2in 2.000 kg of H2O. • Find the molality of 180.0 g of H2O in 1.500 kg of ethyl alcohol. 1.667 m 0.8585 m 1.667 m 6.667 m

9. Molality Example 2: How many grams of potassium iodide, KI, M = 166.0 g/mol, must be dissolved in 500.0 g of water to produce a 0.060 molalsolution? msoln = 0.060 mol/kg mH2O = 0.500 kg MKI = 166.0 g/mol msoln = nKI/mH2O mKI = MKInKI First, we write down the known values. To calculate mKI, the mass of KI, we need to know the number of mols of KI, nKI. We only know the molality of the solution, msoln. We can rearrange the equation to solve for nKI… … we substitute in known values … … and do the calculation. Now, we can calculate mKI, the mass of KI needed to prepare the solution. We put in the known values … … and do the calculation. nKI = msolnmH20 nKI = (0.060 mol/kg)(0.500 kg) = 0.030 mol = (166.0 g/mol)(0.030 mol) = 5.0 g

10. Molality Example Problems: • Find the mass of NaCl (58.5 g/mol) needed to prepare a 0.600 m solution with a mass of 2.50 kg of water. • Find the mass of NaOH (40.0 g/mol) needed to prepare a 1.20 msolutionwitha mass of 0.250 kg of water. • Find the mass of CrF2 (90.0 g/mol) needed to prepare a 1.50 msolution with a mass of 0.400 kg of water. • Find the mass of KNO3 (101 g/mol) needed to prepare a 1.01 msolution with a mass of 0.100 kg of water. 87.8 g NaCl 12.0 g NaOH 54.0 g CrF2 10.2 g KNO3

11. Mole Fraction • Mole Fraction is the ratio of number of mols of the solute to the total number of mols of the solute plus the solvent. • We use the symbol Χ (the Greek letter chi) to represent the mole fraction. • Χsolute = nsolute nsolute + nsolvent

12. Mole Fraction Example 3 Ethylene glycol (EG), C2H6O2, is added to automobile cooling systems to protect against cold weather. What is the mole fraction of each component in a solution containing 1.25 mols of EG and 4.00 mols of water? nEG = 1.25 mol nH2O = 4.00 mol ΧEG = ΧH2O= nEG nEG 1.25 mol 4.00 mol = 0.238 = = nEG + nH2O nEG + nH2O (1.25 + 4.00) mol (1.25 + 4.00) mol = 0.762

13. Mole Fraction Example Problems: • Find the mole fraction of 1.00 mol of NaCl in 55.6 mols of water. • Find the mole fraction of 5.00 mol of BaCl2in 10.0 molsof water. • Find the mole fraction of 0.240 molof C6H12O6 in 4.76 mols of ethanol. • Find the mole fraction of 0.0320 molof KNO3in 10.2 mols of water. Χ = 0.177 Χ = 0.333 Χ = 0.0480 Χ = 0.00319

14. Colligative Calculations • The magnitudes of freezing point depression (∆Tf) and boiling point elevation (∆Tb) are • directly proportional to the molal concentration of the solute, • if the solute is molecular and not ionic.

15. Colligative Calculations • The magnitudes of freezing point depression (∆Tf) and boiling point elevation (∆Tb) are • directly proportional to the molal concentration of all solute ions in the solution, • if the solute is ionic.

16. Colligative Calculations • ∆Tf = −Kfx m • where • ∆Tfis the freezing point depression of the solution. • Kfis the molalfreezing point constant for the solvent. • mis the molal concentration of the solution.

17. Colligative Calculations • ∆Tb= Kbx m • where • ∆Tbis the boilingpoint elevationof the solution. • Kbis the molalboilingpoint constant for the solvent. • mis the molal concentration of the solution.

18. Colligative Calculations The number of particles in solution is equal to the number of mols of the compound. This is a non-ionic compound Example 4 What is the freezing point depression, ∆Tf, of a benzene (C6H6, 78.1 g/mol, Bz) solution containing 400 g of Bz and 200 g of the compound acetone (C3H6O, 58.0 g/mol, Ac)? Kf for Bz is 5.12°C/m. mBz= 400 g = 0.400 kgmAc= 200 g MAc = 58.0 g/molKf= 5.12°C/m nAc= msoln= nAc mAc 3.45 mol 200 g = 3.45 mol = = mBz MAc 58.0 g/mol 0.400 kg = 8.62 m ∆Tf = −Kf × m = (5.12°C/m)(8.62 m) = −44.1°C

19. Colligative Calculations For each mol of NaCl in solution, there is 1 mol of Na+ and 1 mol of Cl− ions. We need to multiply n by 2. The number of particles in solution is equal to the number of ions of the compound. This is an ionic compound Example 5 What is the freezing point depression, ∆Tf, of a sodium chloride (NaCl, 58.4 g/mol) solution containing 87.6 g of NaCl in 1.20 kg water? Kf for water is 1.86°C/m. mH2O= 1.20 kgmNaCl= 87.6 g MNaCl = 58.4 g/molKf= 1.86°C/m nNaCl= msoln= = 1.50 molNaCl = 3.00 mol nNaCl mNaCl 3.00 mol 87.6 g = = mH2O MNaCl 58.4 g/mol 1.20 kg = 2.50 m ∆Tf = −Kf × m = (1.86°C/m)(2.50 m) = −4.65°C

20. Colligative Calculations Example Problems: Find the freezing point depression of: • 58. 44 g of NaCl (M = 58.44 g/mol) in 1.00 kg of H2O. (Kf= 1.86°C/m) • 228 g of octane, C8H18(M = 114 g/mol), in 0.500 kg of benzene. (Kf = 5.12°C/m) • 33.5 g of NaC2H3OH (M = 67.0 g/mol) in 1.25 kg of acetic acid. (Kf = 3.90°C/m) • 217 g of C5H12(M = 72.2 g/mol) in 0.600 kg of cyclohexane. (Kf = 20.2°C/m) ∆Tf = −3.72°C ∆Tf= −20.5°C ∆Tf= −3.12°C ∆Tf= −101°C

21. Colligative Calculations Example 6 What is the boiling point, T, of a benzene (Bz) solution containing 0.600 kg of Bz and 200 g of the compound phenol (C6H5OH, 94.1 g/mol, Ph)? Kb for Bz is 2.53°C/m. Normal boiling point for Bz is Tb= 80.1°C. mBz= 0.600 kgmPh= 200 g Tb = 80.1°C MPh = 94.1 g/molKf= 5.12°C/m nPh= msoln= = 2.13 mol nPh mPh 200 g 2.13mol = = mBz MPh 94.1 g/mol 0.600kg = 3.54m ∆Tb = Kb × m = (2.53°C/m)(3.54 m) = 8.96°C T = Tb+ ∆Tb = 80.1°C + 8.96°C = 89.1°C

22. Colligative Calculations Example 7 What is the boiling point, T, of a sodium chloride (NaCl, 58.4 g/mol) solution containing 87.6 g of NaCl in 0.800 kg of acetic acid (AcOH)? Kb for AcOH is 3.07°C/m. The boiling point of AcOH is Tb= 118.5°C. mH2O= 1.20 kgmNaCl= 87.6 g Tb = 118.5°C MNaCl = 58.4 g/mol Kb= 3.07°C/m nNaCl= msoln= = 1.50 molNaCl = 3.00 mol particles mNaCl nNaCl 87.6 g 3.00 mol = 3.75 m = = mAcOH MNaCl 58.4 g/mol 0.800 kg ∆Tb = Kb × m = (3.07°C/m)(3.75m) = 11.5°C T = Tb+ ∆Tb = 118.5°C + 11.5°C = 130.0°C

23. Colligative Calculations Example Problems: Find the boiling point elevation of: • 58. 44 g of NaCl (M = 58.44 g/mol) in 1.00 kg of H2O. (Kb= 0.512°C/m) • 228 g of octane, C8H18(M = 114 g/mol), in 0.500 kg of benzene. (Kb= 2.53°C/m) • 33.5 g of NaC2H3OH (M = 67.0 g/mol) in 1.25 kg of acetic acid. (Kb= 3.07°C/m) • 217 g of C5H12(M = 72.2 g/mol) in 0.600 kg of cyclohexane. (Kb= 2.79°C/m) ∆Tb = 1.02°C ∆Tb= 10.1°C ∆Tb= 2.46°C ∆Tb= 14.0°C

24. Summary • Molality - we measure the number of mols of solute in the mass of solvent. • msolution = nsolute/msolvent • The mass of the solvent is measured in kilograms, kg. • Mole Fraction is the ratio of number of mols of the solute to the total number of mols of the solute plus the solvent. • We use the symbol X to represent the mole fraction. • Xsolute= (nsolute)/(nsolute + nsolvent)

25. Summary • ∆Tf = Kf x m • ∆Tf is the freezing point depression of the solution • Kfis the molal freezing point constant for the solute • mis the molal concentration of the solution. • ∆Tb = Kb x m • ∆Tb is the boiling point elevation of the solution • Kbis the molal boiling point constant for the solute • mis the molal concentration of the solution.