Chapter 12: Solutions Principles of Chemistry A Molecular Approach,1st Ed. Nivaldo Tro NaCl soln Dr. Azra Ghumman Memorial University of Newfoundland Chem 1011 Solution Solution-A homogeneous mixture of two pure substances. Solution = solute + solvent Solute- The solution component in smaller amount Solvent- The solution component in greater amount Most homogeneous materials are actually solutions. e.g. air and seawater (NaCl + H2O) Nature has a tendency toward spontaneous mixing. Generally, uniform mixing is more energetically favorable. Chem 1011 Chapter 12: Solutions A. Ghumman 2 Sea Water is a thirsty solution. Why? Nature’s tendency towards mixing and lower conc. of salt in body fluids causes a flow of solvent out of body cells into the sea water Chem 1011 Chapter 12: Solutions A. Ghumman 3 12.2-Types of solutions Gaseous solution- Gases or vapours are miscible in all proportions. e.g. Air Liquid solution-obtained by dissolving a gas, liquid or solid in water e.g. soda water (gas-liquid), ethanol water (liquidliquid) salt solution (solid-liquid), and ocean water Solid-solid solution- e.g. Ag in Hg, (dental filling), Au-Ag alloy,Jewelry, Brass(Cu&Zn) Chapter 12: Solutions Chem 1011 A. Ghumman 4 Solubility Solubility- The amount of the substance that will dissolve in a given amount of solvent E.g. Solubility of NaCl in water at 25 °C is 36g NaCl/ 100g H2O The solubility of one substance in another depends on two factors nature’s tendency towards mixing(Entropy), the types of IMF’s The solubility varies with temperature and pressure. Miscible- Two liquids that are mutually soluble are said to be miscible Chem 1011 Chapter 12: Solutions A. Ghumman 5 Mixing and the Solution Process Entropy Most processes occur because the end result has less potential energy However, a solution formation does not necessarily lower the potential energy of the system Entropy-Is a measure of energy randomization or energy dispersal in a system When two ideal gases are put into the same container, they mix spontaneously. Why? Chem 1011 Chapter 12: Solutions A. Ghumman 6 Solution formation Process of solution formation depends on IMF’s between solute and solvent molecules 1. solute–solute attractive forces 2. solvent–solvent attractive forces Both processes are endothermic. 3. solute–solvent attractions. (exothermic) A solution will form if attractive forces between solute and solvent molecules are stronger than between pure solvent molecules and pure solute molecules, A rule of thumb is “like dissolve like” Chem 1011 Chem 1011 Chapter 12: Solutions A. Ghumman Chapter 12: Solutions A. Ghumman 7 8 Common Laboratory solvents Chem 1011 Chapter 12: Solutions A. Ghumman 9 Practice Problem Determine whether each of the following compound is soluble in hexane (C6H6) a) H2O b) Propane(CH3CH2CH3) c) Ammonia (NH3) d) Hydrogen chloride (HCl) Chem 1011 Chapter 12: Solutions A. Ghumman 10 Describing a Solution Qualitative description Concentrated solution-contains large quantities of solute relative to amount of solvent Dilute solution-contains small quantity of solute compared to solvent Saturated A solution that has the maximum amount of solute dissolved in it Supersaturated-A solution that has more solute than saturation Chem 1011 Chapter 12: Solutions A. Ghumman 11 12.5- Expressing Solution Concentration Quantitative description Concentration- The amount of solute in given amount of solvent or solution Molarity(M)- Amount of solute in moles divided by the volume of solution in liters (mol L-1) Molarity is temperature dependent Percent by mass or mass %- Chem 1011 Chapter 12: Solutions A. Ghumman 12 Solution Concentration Chem 1011 Chapter 12: Solutions A. Ghumman 13 Solution Concentrations Mole Fraction, XA The mole fraction- The fraction of the moles of one component in the total moles of all the components of the solution Sum of all the mole fractions in a solution = 1 unitless The mole percentage: mol% = XA × 100% Chem 1011 Chapter 12: Solutions A. Ghumman 14 Practice problems 1. A solution prepared by mixing 17.2 g of ethylene glycol(Mm C2H6O2 = 62.07g mol-1 ) with 0.500 kg of H2O to make 515 mL of solution. Calculate its a) molarity, b) % by mass c) mol fraction d) mol % b) 2. How would you prepare 250.0 mL of 19.5% by mass CaCl2 solution? (d = 1.18 g/mL) Chem 1011 Chapter 12: Solutions A. Ghumman 15 12.6 Colligative Properties Colligative properties-Properties whose value depends only on the number of solute particles, and not on the type of particles Consequently, ionic solutes that dissociate in solution provide higher effective solute concentration than nonelectrolytes. Examples include vapour pressure lowering of a volatile solvent by the addition of nonvolatile solute, freezing point depression (antifreeze, 0.01mol of Ethyleneglycol lower the of H2O) boiling point elevation osmotic pressure Chapter 12: Solutions A. Ghumman Chem 1011 16 Osmosis and Osmotic pressure Osmosis-The flow of solvent from a solution of low concentration into a solution of high concentration through a semipermeable membrane. A semipermeable membrane allows solvent molecules to flow through it, but not the solute osmotic pressure-The amount of pressure needed to stop osmotic flow from taking place MRT M is the molar concentration in mol L-1 R, the gas constant = 0.0821 L.atm mol-1.K-1 T is Temperature in Kelvin Chem 1011 Chapter 12: Solutions A. Ghumman 17 Osmotic Pressure An osmosis cell- Chem 1011 Chapter 12: Solutions A. Ghumman 18 Reverse Osmosis In normal osmosis solvent flows from a dilute solution through a membrane to a more concentrated solution Reverse Osmosis- is reversing the osmosis process by applying greater pressure than osmotic pressure to the more concentrated solution so that solvent flows from concentrated solution through a membrane to the more dilute solution Desalination- removing salts from seawater to make it usable for drinking Chapter 12: Solutions A. Ghumman Chem 1011 19 Calculating molar mass What is the molar mass of a protein if 5.87 mg per 10 mL solution gives an osmotic pressure of 2.45 torr at 25°C? Given: mass of protein 5.87 mg Vsoln = 0.001L = 2.45 torr T = 25 °C + 273.15 = 298.15 K Find: molar mass, g/mol 1. we know we can calculate M 2. Calculate moles of protein using n = MV 3. Calculate molar mass Chem 1011 Mm = mass (g)/mol Chapter 12: Solutions A. Ghumman 20 Importance of Osmosis Rising of Sap in a tree trunk- The sap flows upward to dilute the water solution in the leaves during the day Osmosis is important in many biological processes The solution surrounding the cell must have osmotic pressure equal (isotonic) to that within the cell, why? For intravenous feeding it is necessary that the nutrient solution have exactly the osmotic pressure of blood plasma, otherwise the cell may collapse or burst as a result of osmosis Chem 1011 Chapter 12: Solutions A. Ghumman 21 The Importance of Osmotic Pressure to Living cells Osmosis in a blood cell Osmotic pressure of the solution is greater than that of the cell (cell collapsed). Hypertonic solution Chem 1011 Osmotic pressure of the solution is equal to that of the cell (normal cell) Isotonic Chapter 12: Solutions A. Ghumman Osmotic pressure of the solution is less than that of the cell (bloated shape) Hypotonic 22 This document was created with Win2PDF available at http://www.win2pdf.com. The unregistered version of Win2PDF is for evaluation or non-commercial use only. This page will not be added after purchasing Win2PDF.
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