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**Calculations & Colligative Properties**

Chapter 16.4

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**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

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**Molarity vs Molality Molarity (M) moles of solute M =**

liters of solution Molality (m) m = moles of solute mass of solvent (kg)

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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

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**Why do we need to know molality?**

Colligative properties ….. We use molality to make calculations for boiling point elevation or freezing point depression

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**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)

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**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)

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**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

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**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

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Mole Fraction Mole fractions have no units XA XB = 1

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**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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1 CHAPTER 14 Solutions The Dissolution Process 1.Effect of Temperature on Solubility 2.Molality and Mole Fraction Colligative Properties of Solutions 3.Lowering.

1 CHAPTER 14 Solutions The Dissolution Process 1.Effect of Temperature on Solubility 2.Molality and Mole Fraction Colligative Properties of Solutions 3.Lowering.

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