FSU CHM 1046 - Exam 2 Study Guide
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Exam 2 Study Guide Chapter 12 12 3 Know the different concentration units Be able to use them in any calculation I Concentration The amount of solute present in a given quantity of solvent or solution a by Mass i by mass ii Example mass of solute mass of solute solvent 100 mass of solute mass of solution 100 1 A sample of 892g of KCl is dissolved in 54 6g of H2O What is the by mass of KCl a mass of solute mass o f solute solvent 100 b 892 g 54 6 g 892 g c by mass 1 61 100 b Mole Fraction i X A ii X A Moles of A of moles of all components Moles of A of moles of all components iii Example 100 Percentage of mole fraction 1 3 moles of H2 1 mole of N2 with 2 moles of O2 Calculate the mole fraction X a X A Moles of A of moles of all components b of mole s of all components 1 2 3 6 c X H 2 d X N 2 e X O 2 1 2 Moles of H 2 6 Moles of N 2 6 Moles of O 2 6 3 6 1 6 2 6 1 3 c Molarity M i M moles of solute Liters of solutio n ii Side note molarity is affected by temperature and is therefore less accurate than molality d Molality m i m moles of solute mass of solvent kg Make sure your mass of solvent is in kg and NOT g ii We can use molarity to calculate molality and vice versa iii Example 1 Calculate the molality of a solution made by mixing 1 45g of glucose C6H12O6 and 35g of H2O 008 mol of glucose a Convert 1 45 g glucose to moles 1 mole of glucose 180 16 g i 1 45g b Convert 35g of H2O to kg 35 g H 2O 1000 i 035 kg H 2 O c Solve i m ii m moles of solute mass of solvent kg 008 molglucose 035 kg H 2 O iii m 229 II We can use molarity to calculate molality and vice versa a Example i We have 396m glucose MW 180 16 g in H2O Assume that the density of the solution is 1 16g ml Calculate the molarity of the solution 1 Formulas and knowledge we need a m b M moles of glucose mass of H 20 kg Moles of glucose Volume of solution d V mass of solution Density of solution c mass of solution mass of solute mass of solvent e We must assume we have 1 kg of solvent whenits massis not stated So 1 kg of H2O 2 Now Solve a m moles of glucose mass of H 20 kg b moles of glucose X 1 kg 396 mols of glucose 396 m 1 kg c Assuming we have 1000g of H2O 1 kg our solvent and knowing that we have 396 moles of glucose we now need to start solving for Molarity d M Moles of glucose Volume of solution e To find the volume of solution we first need to find the mass of solution To do this we need the mass of our solvent remember mass of solution mass of solvent mass of solution f Mass of glucose 396 mol 180 2 g mol 71 4 g of glucose g Mass of solution 71 4 g 1000 g 1071 4 g solution h We now have the grams of solution We need to relate grams to volume so the formula that does this is Density i D mass volume ii Volume iii Volume mass density 1071 4 g 1 16 g ml iv Volume 923 6 ml v Convert this volume to Liters because the formula for molarity is only in Liters 1 So volume 9236 L i We can finally solve for Molarity 396 mol 9236 L Moles of glucose Volume of solution j M k 429M 12 4 Know the effect of temperature on solubility of solids figure 12 3 Know the effect of temperature on solubility of gases Figure 12 3 12 3 this is not the exact figure from book but it is the same graph representing the same idea This graph is demonstrating the temperature dependence of the solubility of some ionic compounds in water This also shows that there is no standard for the effect of temperature on a solid s solubility Chang 522 I Solubility and Temperature For solids There is no standard for solubility Some will increase decrease or stay the same with an increase decrease in temperature For gases Solubility of a gas will decrease as temperature increases and vice versa In other words solubility of a gas 1 T 12 5 Know the effect of pressure on solubility of gases Understand Henry s law and be able to use it in calculations Pressure on Solubility a Pressure does NOT affect solubility of solids and liquids b Pressure DOES affect solubility of gases I II i As pressure increases solubility of the gas also increases Henry s Law a The solubility of a gas in a liquid is proportional to the pressure of the gas over the solution i c kP b Example i The solubility of N2 g at 25 C and 1 atm equals 6 8x10 4 mol L What is c the concentration M of the dissolved gas P pressure of the gas over the solution K is the constant for each gas at a certain temperature mol L atm the concentration M of nitrogen dissolved in H2O under atmospheric conditions Partial Pressure of N2 78 atm Identify the givens ii 1 CN2 6 8x10 4 mol L this is the solubility of nitrogen ALONE We need to use this 2 Rearrange equation to solve for k value to find k b k a c kP c P 6 8 x 10 4 mol L 1atm 6 8 x 10 4 mol L atm d k c k 3 We must now use this k to solve for c a c kP b c 78 c c 0005 6 8 x 10 4mol L atm 12 6 Colligative Properties of Nonelectrolytes Know what the definition of colligative properties is Know all four colligative properties vapor pressure lowering boiling point elevation freezing point depression and osmotic pressure Understand Raoult s Law and be able to use it in calculations for both volatile and non volatile solutes Understand both Tb and Tf expressions and be able to use them in calculations Know what Osmotic pressure is Know the three types of solutions isotonic hypertonic and hypotonic Understand the expression of osmotic pressure and be able to use it in calculations Be able to use any colligative property to find molar mass I Colligative Properties Properties that depend only on the number of solute particles in solution and not on the nature of the solute particles a Vapor Pressure Lowering i Raoult s Law describes the relationship between vapor pressure and a 1 For nonvolatile solutions AKA the solution only contains ONE solute …

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# FSU CHM 1046 - Exam 2 Study Guide

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