Presentation is loading. Please wait.

Presentation is loading. Please wait.

Solution A homogeneous mixture of two or more substances.

Similar presentations


Presentation on theme: "Solution A homogeneous mixture of two or more substances."— Presentation transcript:

1

2 Solution A homogeneous mixture of two or more substances.

3 Solvent Solvent - the substance present in the most amount. Solute Solute - the substance present in the least amount. For Any Solution

4 Solvent Solvent - there can be only ONE Solute Solute - there can be MORE than one For Any Solution

5 Solvent Solvent - the substance present in the most amount. Solute Solute - the substance present in the least amount. What is the solvent in air? in air?

6 Solvent Solvent - the substance present in the most amount. Solute Solute - the substance present in the least amount. What are the solutes in air? in air?

7 Solvent Solvent - the substance present in the most amount. Solute Solute - the substance present in the least amount. What is the solvent in stainless steel? in stainless steel?

8 Solvent Solvent - the substance present in the most amount. Solute Solute - the substance present in the least amount. What are the solutes in stainless steel? in stainless steel?

9 Solvent Solvent - Water Solute Solute - The substance dissolved in water Aqueous Solution

10 The amount of solute that can be dissolved in a given amount of solvent, at a given temperature. Solubility

11

12

13 Is NaCl soluble in H 2 O?

14 A solution containing the maximum amount of a solute that is possible to dissolve in a given volume of solvent at a given temperature. Saturated Solution

15 NaCl has a solubility of 357 grams per liter of "cold" H 2 O. Saturated Solution +

16 How many moles of NaCl are in the solution? Saturated Solution NaClsolution

17 357 g NaCl 1 mole NaCl Saturated Solution NaClsolution 58 g NaCl 6.2 mole NaCl =

18 How many "particles" of NaCl are in the solution? Saturated Solution NaClsolution

19 357 g NaCl Saturated Solution NaClsolution 58 g NaCl 3.7 X particles NaCl = 6.02 X particles NaCl

20 A comparison of the amounts of solute and solvent in a solution. Concentration

21 "Strong" and "Weak" give SOME comparison, but only a general idea. Concentration

22 "Dilute" and "Concentrated" still don't provide enough for quantitative calculations. Concentration

23 To do calculations, we must know "how much" solute and "how much" solvent are present. Concentration

24 M = Molarity moles solute dm 3 solution

25 1 mole = formula mass (g) 1 liter = 1 cubic decimeter dm 3 1 liter = 1000 milliliters 1 ml = 1 cm 3

26 1. What is the molarity of a liter of solution containing 100 grams of copper (II) chloride? Molarity = moles dm 3

27 1 0 0 g C u C l 2 1 l i t e r Molarity = moles dm 3

28 1 0 0 g C u C l 2 1 l i t e r Molarity = grams moles dm 3

29 1 0 0 g C u C l 2 1 l i t e r Molarity = grams mole moles dm 3

30 1 0 0 g C u C l 2 1 l i t e r Molarity = grams 1 mole moles dm 3

31 1 0 0 g C u C l 2 1 m o l e 1 l i t e r g Molarity = Cu = 1 X 64 = 64 Cl = 2 X 35 = moles dm 3

32 1 0 0 g C u C l 2 1 m o l e 1 l i t e r 1 l i t e r g 1 d m 3 Molarity = moles dm 3

33 1 0 0 g C u C l 2 1 m o l e 1 l i t e r 1 l i t e r g 1 d m 3 Molarity = moles dm 3 Have we worked the problem?

34 1 0 0 g C u C l 2 1 m o l e 1 l i t e r 1 l i t e r g 1 d m M C u C l 2 Molarity = moles dm 3

35 2.How much NaCl is needed to prepare 250ml of 0.5M salt water How is this problem different from the first?

36 1. What is the molarity of a liter of solution containing 100 grams of copper (II) chloride? 2.How much NaCl is needed to prepare 250ml of 0.5M salt water? The second GIVES M The first ASKS for M

37 2.How much NaCl is needed to prepare 250 ml of 0.5 M salt water

38 250 ml 0.5 mole NaCl dm 3 Preparation

39 250 ml 0.5 mole NaCl dm 3 Preparation What units will we have when the problem is worked?

40 250 ml 0.5 mole NaCl dm 3 Preparation need grams

41 250 ml 0.5 mole NaCl dm 3 Preparation need grams

42 250 ml 0.5 mole NaCl dm 3 1 mole NaCl Preparation need grams

43 250 ml 0.5 mole NaCl 58 g NaCl dm 3 1 mole NaCl Preparation need grams

44 250 ml 0.5 mole NaCl 58 g NaCl dm 3 1 mole NaCl Preparation Now What? need grams

45 250 ml 0.5 mole NaCl 58 g NaCl 1 dm 3 dm 3 1 mole NaCl 1000 ml Preparation need grams

46 250 ml 0.5 mole NaCl 58 g NaCl 1 dm 3 dm 3 1 mole NaCl 1000 ml Preparation 7.3 grams NaCl

47 Practice Problems

48 0.37 g CaCl 2 1 mole 1000 ml 340 ml 110 g 1 dm M CaCl 2 Practice Problem 1

49 50 cm mole Al(OH) 3 78 g Al(OH) 3 1 dm 3 dm 3 mole 1000 cm g Al(OH) 3 Practice Problem 2

50 Homework Problems

51 50 g NaOH 1 mole 1000 cm cm 3 40 g 1 dm M NaOH Homework Problem 1

52 100 cm mole CaSO g 1 dm 3 d m 3 mole 1000 ml 3.4 g CaSO 4 Homework Problem 2

53 100 ml 0.5 mole HCl 36 g 1 dm 3 dm 3 mole 1000 ml 1.8 g HCl Homework Problem 3

54 Making Dilutions

55 A solution can be made less concentrated by dilution with solvent Making Dilutions

56 M 1 V 1 = M 2 V 2 original solution 1 = diluted solution 2 Volume units must be the same for both volumes in this equation. Making Dilutions

57 How do you prepare 100ml of 0.40M MgSO 4 from a stock solution of 2.0M MgSO 4 ? Dilution Problem

58 M 1 V 1 = M 2 V 2 M 1 = V 1 = M 2 = V 2 = Dilution Problem

59 M 1 V 1 = M 2 V 2 M 1 = 2.0M MgSO 4 V 1 = unknown M 2 = 0.40M MgSO 4 V 2 = 100ml Dilution Problem

60 Step 1 - write the equation: M 1 V 1 = M 2 V 2 Step 2 - manipulate the equation: V 1 = M 2 V 2 / M 1 Step 3 - add the numbers: V 1 = (0.40M) (100ml) / 2.0M Dilution Problem

61 Step 4 - do the calculation: V 1 = 20ml Step 5 - describe the preparation: Add 80ml of distilled water to 20ml of the 2.0 M MgSO 4 solution Dilution Problem

62 Homework Problems

63 Molarity Calculations: M (NH 4 ) 2 C 4 H 4 O M CoSO M Fe(NO 3 ) 2

64 Preparations: 1.390g NiCl g AgF 3.Use 50cm 3 of the 1.0M NaCl solution. Add 200cm 3 of distilled water to make the total volume 250cm 3.

65 Dilutions: 1.Add 19.2cm 3 of 0.52M CoCl 2 solution to a graduate. Add distilled water to make the total volume 100cm 3. 2.Add 72.5cm 3 of 0.69M Ba(NO 3 ) 2 solution to a graduate. Add distilled water to make the total volume 200cm 3. 3.Add 37.5ml of 2M NH 4 Br solution to a graduate. Add distilled water to make the total volume 500ml.

66 Solution Preparation

67 Other Solution Concentrations

68 Molality moles solute m = Kg solvent

69 Normality equivalents equivalents solute N = dm 3 solution

70


Download ppt "Solution A homogeneous mixture of two or more substances."

Similar presentations


Ads by Google