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Exam 2 Study GuideChapter 1212.3• 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 = mass of solutemass of solute +solvent×100=mass of solutemass of solution×100ii. Example: 1. A sample of .892g of KCl is dissolved in 54.6g of H2O. What is the% by mass of KCl? a.mass of solutemass o f solute+solvent× 100b..892 g54.6 g+.892 g× 100c.% by mass=1.61 %b. Mole Fraction i.XA=Moles of A∑of moles of all componentsii.XA=Moles of A∑of moles of all components×100 = Percentage of mole fractioniii. Example: 1. 3 moles of H2, 1 mole of N2, with 2 moles of O2. Calculate the molefraction (X). a.XA=Moles of A∑of moles of all componentsb.∑of mole s of all components=1+2+3=6c.XH 2=Moles of H 26=36=12d.XN 2=Moles of N 26=16e.XO 2=Moles of O 26=26=13c. Molarity (M) i. M = molesof soluteLitersof solutio nii. Side note: molarity is affected by temperature and is therefore less accurate than molality. d. Molality (m) i.m=moles of solutemass 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. a. Convert 1.45 g glucose to molesi. 1.45g ×1 mole of glucose180.16 g=.008 mol of glucoseb. Convert 35g of H2O to kgi.35 g H 2O1000=.035 kg H 2 Oc. Solve: i.m=moles of solutemass of solvent(¿kg)ii.m=.008 molglucose.035 kg H 2 Oiii.m=.229II. 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 glucosemass of H 20(¿kg)b.M=Moles of glucoseVolume of solutionc.mass of solution=mass of solute+mass of solventd.V =mass of solutionDensity of solutione.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 glucosemass of H 20(¿kg) b.moles of glucose=.396 m1(kg)X 1 kg=.396 mols of glucosec. Assuming we have 1000g of H2O (1 kg), our solvent, and knowing that we have .396 moles of glucose, we now needto start solving for Molarity.d.M=Moles of glucoseVolume of solutione. 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 gmol)=71.4 g of glucoseg. Mass of solution = 71.4 g + 1000 g = 1071.4 g solutionh. We now have the grams of solution. We need to relate grams to volume so the formula that does this is… Density! i. D= massvolumeii. Volume= massdensityiii. Volume = 1071.4 g1.16 g /mliv. Volume= 923.6 ml v. Convert this volume to Liters because the formula for molarity is only in Liters. 1. So volume = .9236 Li. We can finally solve for Molarity!j.M=Moles of glucoseVolume of solution=..396 mol.9236 Lk. = .429M12.4 • 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 α 1T12.5• Know the effect of pressure on solubility of gases.• Understand Henry’s law and be able to use it in calculationsI. Pressure on Solubility:a. Pressure does NOT affect solubility of solids and liquidsb. Pressure DOES affect solubility of gases. i. As pressure increases,solubility of the gas alsoincreases. II. Henry’s Lawa. The solubility of a gas in a liquidis proportional to the pressure ofthe gas over the solution. i. c = kP b. Example i. The solubility of N2(g) at25°C and 1 atm equals6.8x10-4 mol/L. What isc = the concentration (M) of the dissolved gas P= pressure of the gas over the solutionK is the constant for each gas at a certain temperature (molL∗atm)the concentration (M) of nitrogen dissolved in H2O under atmospheric conditions? Partial Pressure of N2 = .78 atm ii. Identify the givens!1. CN2= 6.8x10-4 mol/L this is the solubility of nitrogenALONE. We need to use this value to find k. 2. Rearrange equation to solve for ka. c = kPb. k = cPc.k=6.8 x 10−4mol / L1atmd.k =6.8 x 10−4molL∗atm3. We must now use this k to solve for c. a. c = kP b. c = .78(6.8 x 10− 4molL∗atm¿c. c = .000512.6 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 ΔTb and ΔTf 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 = X1P10i. P1 = Vapor Pressure of Solutionii. 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 solute (X2) = 1 iii. P10= vapor pressure of pure solvent b. Vapor Pressure lowering: i.∆ P=P10−P1c. Example: i. Calculate the vapor pressure of a solution made by dissolving 218 g of glucose (molar mass = 180.2 g/mol) in 460 mL of water at 30°C.ii. What

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