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Calculations & Colligative Properties

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Presentation on theme: "Calculations & Colligative Properties"— Presentation transcript:

1 Calculations & Colligative Properties
Chapter 16.4

2 Learning Objectives 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

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

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

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

6 Boiling Point Elevation
DTb = Kb x m x i Van’t Hoff factor Change in boiling point Molality of solution Molal boiling point constant (0.51 oC/m for water)

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

8 Important note about these calculations!
The calculations change depending on whether you have a nonelectrolyte or electrolyte solution Why? 1 C12H22O11 (s)  1 C12H22O11 (aq) (i =1) Sugar does not dissociate into ions! 1 Ba(NO3)2 (s)  1 Ba2+(aq) 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 Sample Calculation: Freezing Point Depression: Electrolyte
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? DTf = Kf •m•i 137.3(1) 14(2) 16(6)_____ 261.3g/mol 800. g = kg Let’s calculate m first m = moles of Ba(NO3)2 / kg solvent m = moles / kg water 62.5 g 261.3 g/mol m = mol/kg DTf = (-1.86 oC/m)(0.299 m)(3)  Van’t Hoff factor (i) DTf = 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 = 1.43 / XA = 0.293 XA = 3.45 / XA = 0.707 = 1

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