1 / 18

Calculations Involving Colligative Properties

Calculations Involving Colligative Properties. 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. Introduction.

edan-knight
Télécharger la présentation

Calculations Involving Colligative Properties

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Calculations Involving Colligative Properties

  2. 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

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

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

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

  6. Molality Example 1 Find the molality of 87.66 g of sodium chloride dissolved in 2.500 kg of water. mNaCl = 87.66 g mH2O = 2.500 kg MNaCl = 58.44 g/mol m = nNaCl/mH2O = 1.500 mol/2.500 kg m = 0.600 mol/kg nNaCl = mNaCl/MNaCl = 87.66 g/58.44 g/mol nNaCl = 1.500 mol

  7. Molality Example 2 How many grams of potassium iodide must be dissolved in 500.0 g of water to produce a 0.060 molal KI solution? msolution = 0.060 m mH2O = 500.0 g = 0.5000 kg MKI = 166.0 g/mol m = nKI/mH2O nKI = m x mH2O = (0.060)(0.5000) mol nKI = 0.030 mol mKI = nKI x MKI = (0.030)(166.0) g = 5.0 g

  8. 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 X to represent the mole fraction. • Xsolute = nsolute nsolute + nsolvent

  9. Mole Fraction Example 3 Ethylene glycol, 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 ethylene glycol (EG) and 4.00 mol of water? nEG = 1.25 mol nH2O = 4.00 mol nEG 1.25 mol XEG = = = 0.238 nEG + nH2O 1.25 mol + 4.00 mol nH2O 4.00 mol XH2O = = = 0.762 nEG + nH2O 1.25 mol + 4.00 mol

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

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

  12. Colligative Calculations • ∆Tf = Kf x m • where • ∆Tf is the freezing point depression of the solution • Kf is the molal freezing point constant for the solvent. • m is the molal concentration of the solution

  13. Colligative Calculations • ∆Tb = Kb x m • where • ∆Tb is the boiling point elevation of the solution • Kb is the molal boiling point constant for the solvent. • m is the molal concentration of the solution

  14. Colligative Calculations Example 4 What is the freezing point depression of a benzene (C6H6, BZ) solution containing 400 g of benzene and 200 g of the compound acetone (C3H6O, AC). Kf for benzene is 5.12°C/m. mAC 200 g mBZ = 400 g = 0.400 kg mAC = 200 g MAC = 58.0 g/mol Kf = 5.12°C/m nAC = = MAC 58.0 g/mol nAC = 3.45 mol nAC 3.45 mol m = = = 8.62 m mBZ 0.400 kg ∆Tf = Kf x m = = (5.12°C/m)(8.62 m) 44.1°C

  15. Colligative Calculations Example 5 What is the boiling point of a 1.50 m NaCl solution? m = 1.50 m Kb = 0.512°C/m Tb = 100.0°C NaCl produces 2 mols of particles for each mol of salt added; m = 2(1.50 m) = 3.00 m ∆Tb = Kb x m = = (0.512°C/m)(3.00 m) 1.54°C T = Tb + ∆Tb = = 100.0°C + 1.54°C 101.5°C

  16. 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.

  17. Summary • 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

  18. Summary • ∆Tf = Kf x m; ∆Tf is the freezing point depression of the solution, Kf is the molal freezing point constant for the solute, and m is the molal concentration of the solution. • ∆Tb = Kb x m; ∆Tb is the boiling point elevation of the solution, Kb is the molal boiling point constant for the solute, and m is the molal concentration of the solution.

More Related