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 ii Example i by mass mass of solute solvent mass of solute 100 massof 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 mass of solute mass of solute solvent 100 a b c 892 g 54 6 g 892 g 100 by mass 1 61 b Mole Fraction X A X A i ii iii Example Moles of A of molesof all components Moles of A of molesof all components 100 Percentage of mole fraction 1 3 moles of H2 1 mole of N2 with 2 moles of O2 Calculate the mole fraction X X A a Moles of A of molesof all components b of mole sof all components 1 2 3 6 X H2 Molesof H2 3 6 1 2 X N2 Moles of N2 1 6 X O2 Moles of O2 2 6 1 3 6 6 6 c d e c Molarity M i M molesof solute Liters of solutio n ii Side note molarity is affected by temperature and is therefore less accurate than molality d Molality m g iii Example m moles of solute massof solvent kg i Make sure your mass of solvent is in kg and NOT ii We can use molarity to calculate molality and vice versa 1 Calculate the molality of a solution made by mixing 1 45g of glucose C6H12O6 and 35g of H2O a Convert 1 45 g glucose to moles i 1 45g 1moleof glucose 180 16 g 008mol of glucose b Convert 35g of H2O to kg 35g H 2O 1000 035 kg H 2O c Solve i i ii m moles of solute mass of solvent kg m 008molglucose 035 kgH2O 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 moles of glucose mass of H20 kg b M Moles of glucose Volume of solution c mass of solution mass of solute mass of solvent V mass of solution Density of solution d e Wemust assume we have 1kgof solvent whenits massis not stated So 1 kg of H2O 2 Now Solve a m moles of glucose mass of H20 kg b molesof glucose 396 m 1 kg X 1kg 396 molsof glucose 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 2g mol 71 4 gof 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 v Convert this volume to Liters because the formula ii Volume mass density iii Volume 1071 4g 1 16g ml iv Volume 923 6 ml for molarity is only in Liters 1 So volume 9236 L i We can finally solve for Molarity M Molesof glucose Volume of solution 396 mol 9236 L j k 429M 12 4 Figure 12 3 Know the effect of temperature on solubility of solids figure 12 3 Know the effect of temperature on solubility of gases 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 I Pressure on Solubility a Pressure does NOT affect solubility of solids and liquids b Pressure DOES affect solubility of gases i As pressure increases solubility of the gas also increases The solubility of a gas in a liquid is proportional to the pressure of the gas over the solution II Henry s Law i c kP b Example 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 i The solubility of N2 g at 25 C and 1 atm equals 6 8x10 4 mol L What is the concentration M of nitrogen dissolved in H2O under atmospheric conditions Partial Pressure of N2 78 atm temperature mol L atm ii Identify the givens 1 CN2 6 8x10 4 mol L this is the solubility of nitrogen ALONE We need to use this value to find k 2 Rearrange equation to solve for k a c kP b k c P k 6 8x 10 4 mol L 1 atm k 6 8x 10 4 mol L atm c d 3 We must now use this k to solve for c a c kP b c 78 6 8x 10 4 mol L atm c c 0005 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 solute 1 For nonvolatile solutions AKA the solution only contains ONE solute a P1 X1P1 0 i P1 Vapor Pressure of Solution ii X1 mole fraction of the solvent 1 Recall that mole fractions of a reaction must equal 1 2 Therefore mole fraction of solvent X1 mole fraction of …
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