Solutions Powerpoint

Definitions ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

Dissolution of Solid Solute ,[object Object],[object Object],[object Object]

Solution and Concentration ,[object Object],[object Object],[object Object],[object Object],[object Object],* Note that molality is the only concentration unit in which denominator contains only solvent information rather than solution.

[object Object],[object Object],[object Object],% Concentration

% Concentration: % Mass Example ,[object Object],[object Object],[object Object],%m = 3.5 g CoCl 2 100g H 2 O = 3.5% (m/m)

Concentration: Molarity Example ,[object Object],Now that the number of moles of substance is known, this can be combined with the volume of solution — which must be in liters — to give the molarity. Because 250. mL is equivalent to 0.250 L . As is almost always the case, the first step is to convert the mass of material to moles. 0.435 g KMnO 4 • 1 mol KMnO 4 = 0.00275 mol KMnO 4 158.0 g KMnO 4 Molarity KMnO 4 = 0.00275 mol KMnO 4 = 0.0110 M 0.250 L solution

[object Object],[object Object],[object Object],[object Object],Dilution A bottle of 0.500 M standard sucrose stock solution is in the lab. Give precise instructions to your assistant on how to use the stock solution to prepare 250.0 mL of a 0.348 M sucrose solution. Suppose you have 0.500 M sucrose stock solution. How do you prepare 250 mL of 0.348 M sucrose solution ? Concentration 0.500 M Sucrose 250 mL of 0.348 M sucrose

3 Stages of Solution Process ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

Dissolution at the molecular level? ,[object Object]

Factors Affecting Solubility 1. Nature of Solute / Solvent . - Like dissolves like (IMF) 2. Temperature - i) Solids/Liquids- Solubility increases with Temperature Increase K.E. increases motion and collision between solute / solvent. ii) gas - Solubility decreases with Temperature Increase K.E. result in gas escaping to atmosphere. 3. Pressure Factor - i) Solids/Liquids - Very little effect Solids and Liquids are already close together, extra pressure will not increase solubility. ii) gas - Solubility increases with Pressure. Increase pressure squeezes gas solute into solvent.

Solubilities of Solids vs Temperature ,[object Object],[object Object]

Temperature & the Solubility of Gases The solubility of gases DECREASES at higher temperatures

Henry’s Law The effect of partial pressure on solubility of gases ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

Henry’s Law & Soft Drinks ,[object Object],[object Object],[object Object],[object Object]

Henry’s Law Application ,[object Object],Step 1: Use the first set of data to find “k” for N 2 at 25 °C Step 2: Use this constant to find the solubility (concentration) when P is 0.78 atm:

Colligative Properties ,[object Object],[object Object],[object Object]

Vapor Pressure Lowering for a Solution ,[object Object],[object Object],[object Object]

Vapor Pressure Lowering ,[object Object]

Application of Vapor Pressure Lowering ,[object Object],[object Object]

Raoult’s Law Describes vapor pressure lowering mathematically. ,[object Object],[object Object],[object Object],[object Object],[object Object],only the solvent (A) contributes to the vapour pressure of the solution

What is the vapor pressure of water above a sucrose (MW=342.3 g/mol) solution prepared by dissolving 158.0 g of sucrose in 641.6 g of water at 25 ºC? The vapor pressure of pure water at 25 ºC is 23.76 mmHg . mol sucrose = (158.0 g)/(342.3 g/mol) = 0.462 mol mol water = (641.6 g)/(18 g/mol) = 35.6 mol P sol’n = X water P  water = (0.987)(23.76 mm Hg) = 23.5 mm Hg

Mixtures of Volatile Liquids Both liquids evaporate & contribute to the vapor pressure

Raoult’s Law: Mixing Two Volatile Liquids ,[object Object],[object Object],[object Object],[object Object],[object Object]

Benzene and Toluene ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

Normal Boiling Process ,[object Object],[object Object],[object Object],[object Object],Elevation of B. pt . Express by Boiling point Elevation equation

Boiling Point Elevation ,[object Object],[object Object],[object Object],[object Object],[object Object],Some Boiling Point Elevation and Freezing Point Depression Constants Normal bp (°C) K b Normal fp (°C) K f Solvent pure solvent (°C/m) pure solvent (°C/m) Water 100.00 +0.5121 0.0 1.86 Benzene 80.10 +2.53 5.50 4.90 Camphor 207 +5.611 179.75 39.7 Chloroform 61.70 +3.63 - 63.5 4.70 (CH 3 Cl)

Freezing Point Depression ,[object Object],[object Object],[object Object],When solution freezes the solid form is almost always pure. Solute particles does not fit into the crystal lattice of the solvent because of the differences in size. The solute essentially remains in solution and blocks other solvent from fitting into the crystal lattice during the freezing process.

Freezing Point Depression ,[object Object], T f = i ·m ·k f Where,  T f = FP depression i = van’t Hoff Factor m = molality , k f = FP Constant Generally freezing point depression is used to determine the molar mass of an unknown substance . Derive an equation to find molar mass from the equation above.

Osmotic pressure ,[object Object],[object Object], = i CRT where P = osmotic pressure i = van’t Hoff factor C = molarity R = ideal gas constant T = Kelvin temperature

Osmosis and Blood Cells (a) A cell placed in an isotonic solution. The net movement of water in and out of the cell is zero because the concentration of solutes inside and outside the cell is the same. (b) In a hypertonic solution, the concentration of solutes outside the cell is greater than that inside. There is a net flow of water out of the cell, causing the cell to dehydrate, shrink, and perhaps die. (c) In a hypotonic solution, the concentration of solutes outside of the cell is less than that inside. There is a net flow of water into the cell, causing the cell to swell and perhaps to burst.

Solutions Powerpoint

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