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Calculations & Colligative Properties Chapter 16.4.

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Presentation on theme: "Calculations & Colligative Properties Chapter 16.4."— 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 M = moles of solute liters of solution Molarity (M) 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 T b = K b x m x i Change in boiling point Molality of solution Vant Hoff factor Molal boiling point constant (0.51 o C/ m for water)

7 Freezing Point Depression T f = K f x m x i Change in freezing point Molal freezing point constant (-1.86 o C/ m for water) Molality of solution Vant Hoff factor

8 Important note about these calculations! The calculations change depending on whether you have a nonelectrolyte or electrolyte solution Why? 1 C 12 H 22 O 11 (s) 1 C 12 H 22 O 11 (aq) ( i =1) Sugar does not dissociate into ions! 1 Ba(NO 3 ) 2 (s) 1 Ba 2+ (aq) + 2 NO 3 -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(NO 3 ) 2, in 800. g of water? T f = K f m i Lets calculate m first m = moles of Ba(NO 3 ) 2 / kg solvent m = moles / kg water 800. g = kg 62.5 g g/mol 137.3(1) 14(2) 16(6)_____ 261.3g/mol m = mol/kg T f = (-1.86 o C/ m )(0.299 m )(3) Vant Hoff factor ( i ) T f = o C

10 Mole Fraction Mole fractions have no units X A + X B = 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? X A = 1.43 / X A = X A = 3.45 / X A = = 1


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