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Calculations & Colligative Properties

Calculations & Colligative Properties. Chapter 16.4. Know the difference between molality and molarity Be able to calculate molality Can use molality to calculate freezing point depression or boiling point elevation Understand how ionic compounds affect colligative properties

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Calculations & Colligative Properties

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  1. Calculations & Colligative Properties Chapter 16.4

  2. Know the difference between molality and molarity • Be able to calculate molality • Can use molality to calculate freezing point depression or boiling point elevation • Understand how ionic compounds affect colligative properties • Can calculate mole fractions Learning Objectives

  3. moles of solute liters of solution moles of solute m = mass of solvent (kg) M = Molarity vs Molality Molarity(M) Molality(m)

  4. so 3.2 M means we have 3.2 moles in 1 liter of solution Molarity vs Molality so 3.2 m means we have 3.2 moles in 1 kg of solvent

  5. Colligative properties ….. Why do we need to know molality? We use molality to make calculations for boiling point elevation or freezing point depression

  6. DTb = Kbxmx i Boiling Point Elevation Van’t Hoff factor Change in boiling point Molality of solution Molal boiling point constant (0.51 oC/mfor water)

  7. DTf = Kfxmx i Freezing Point Depression Van’t Hoff factor Change in freezing point Molality of solution Molal freezing point constant (-1.86 oC/mfor water)

  8. The calculations change depending on whether you have a nonelectrolyte or electrolyte solution Important note about these calculations! Why? 1 C12H22O11 (s)  1 C12H22O11 (aq) (i =1) Sugar does not dissociate into ions! 1 Ba(NO3)2 (s)  1 Ba2+(aq) + 2 NO3-1(aq) (i =3) Barium nitrate will lower the freezing point of its solvent three times as much as sugar of the same molality

  9. What is the freezing point depression of water in a solution of 62.5 g of barium nitrate, Ba(NO3)2, in 800. g of water? Sample Calculation: Freezing Point Depression: Electrolyte 137.3(1) 14(2) 16(6)_____ 261.3g/mol DTf = Kf •m•i 800. g = 0.800 kg Let’s calculate m first m = moles of Ba(NO3)2 / kg solvent m = 0.239 moles / 0.800 kg water 62.5 g 261.3 g/mol m = 0.299 mol/kg DTf = (-1.86 oC/m)(0.299 m)(3)  Van’t Hoff factor (i) DTf = -1.70 oC

  10. Mole Fraction • Mole fractions have no units XA + XB = 1

  11. Mole Fraction Calculation A solution has 1.43 moles of vanadium oxide in 3.45 moles of chloroform. What is the mole fraction of each component in the solution? XA = 3.45 / 1.43 + 3.45 XA = 0.707 XA = 1.43 / 1.43 + 3.45 XA = 0.293 0.293 + 0.707 = 1

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