Aim # 44: How do we calculate the concentration of a solution? H.W. # 44 Study pp. 511 -514 (sec. 11.1) Ans. ques. p. 544 # 29,32-35

I A solution is a homogeneous mixture e. g I A solution is a homogeneous mixture e.g. salt water, ethanol + water, acetone + diethyl ether, brass (Cu + Zn), bronze (Cu + Sn) II Solution components solvent- the component present in the greatest amount solutes)- all other components III Types of solutions A. gaseous- two or more gases mixed uniformly e.g. air SO2, O2, and SO3 N2, H2, and NH3

B. Liquid- a liquid solvent (e. g. water) with 1. a gaseous solute (e B. Liquid- a liquid solvent (e.g. water) with 1. a gaseous solute (e.g. CO2) 2. a liquid solute (e.g. ethanol, bromine) 3. a solid solute (e.g. a salt, sucrose) C. Solid- a solid solvent with 1. a gaseous solute (e.g. H2 in Pd) 2. a liquid solute (e.g. Hg in Ag) 3. a solid solute (e.g. steel, Ag in Au) IV Solution concentrations- qualitative Dilute- a relatively small amount of solute Concentrated- a relatively large amount of solute

Solution concentrations- quantitative A. 1 Solution concentrations- quantitative A. 1. mass percent mass percent = mass solute x 100% mass solution Problem: What is the mass % of sodium chloride in a solution formed by dissolving 25.0 g of NaCl in 0.500 g of H2O? Ans: mass% NaCl = 25.0 x 100% = 4.76% 25.0g + 500.g

2. parts per million (ppm) ppm = mass solute x 106 mass solution (For the above solution, ppm = .0476 x 106 = 4.76 x 104 ppm) Problem: Commercial bleach contains 3.26% sodium hypochlorite, NaOCl by mass. What is the mass of NaOCl in a bottle containing 2500. g of bleach? Ans: 3.26% = mass NaOCl x 100 2500. g mass NaOCl = 3.26 x 2500. = 81.5 g NaOCl 100

B. mole fraction (χ) mole fraction of = χA = moles of component component A total moles of all components Problem: What is the mole fraction of HCl in a solution of 3.65 g HCl dissolved in 90.0 g H2O? Ans: nHCl = 3.65 g HCl x 1 mol HCl = 0.100 mol HCl 36.5 g HCl nH2O = 90.0 g H2O x 1 mol H2O = 5.00 mol H2O 18.0 g H2O χHCl = nHCl = 0.100 mol ͘ nH20 + nHCl 5.00 mol + .100 mol χHCl = 0.100 = .0196 5.10

Ans: mass EtOH = 0.789 g/mL x 10.00 mL = 7.89 EtOH Problem: An ethanol-water solution is prepared by dissolving 10.00 mL of ethanol, CH3CH2OH, (density = 0.789 g/ml) in a sufficient volume of water to produce 100.0 mL of a solution with a density of 0.982 g/ml. What is the mole fraction of the ethanol in the solution? Ans: mass EtOH = 0.789 g/mL x 10.00 mL = 7.89 EtOH moles solution = 7.89 g EtOH = .171 mol EtOH 46.068 g/mol mass solution = 0.982 g/ml x 100.0 mL = 98.2 g solution mass H2O = 98.2 g solution – 7.89 g EtOH = 90.31 g H2O moles H2O = 90.31 g H2O = 5.017 mol H2O 18.0 g/mol χ = nEtOH = .171 mol = .0329 nEtOH + nH2O 5.017 mol + .171 mol

C. Molarity (M) M = molarity = moles solute liters solution What is the molarity of the ethanol solution above? M = nEtOH = .171 mol = 1.71 M Vsol. .100 L D. molality (m) m = molality = moles solute kg solvent Find the molality of the ethanol solution above. Mass H2O = 90.31 g = .09031 kg m = nEtOH = .171 mol = 1.82 m kg H2O .09031 kg

IF YOU KNOW THE MOLARITY OF A SOLUTION YOU CAN CALCULATE THE MOLALITY, MOLE FRACTION, AND MASS % IF YOU ALSO KNOW ITS DENSITY Problem: The molarity of a glycerol solution (C3H8O3 ) solution is .925 M. if the density of the solution is 1.10 g/mL, calculate the molality, mole fraction, and mass % of the solution. Ans: Assume you have 1 L of solution mass solution = 1.10 g/mL x 1000 mL = 1100 g solution moles glycerol = .925 mol/L x 1 L = .925 mol glycerol mass glycerol = .925 mol x 92.09 g/mol = 85.18 g glycerol mass H2O = 1100 g – 85.18 g = 1015 g = .1015 kg H2O m = .925 mol glycerol = 9.11 m .1015 kg H2O

moles H2O = 1015 g x 1 mol = 56.339 mol 18.016 g χ = .925 mol = .0162 .925 mol + 56.339 mol mass % = 85.18 g glycerol x 100 = 77.4% 1100.g Problem: Given that the density of a solution of 5.0 g toluene (C7H8) and 225 g of benzene (C6H6) is 0.876 g/mL, calculate a) the molarity of the solution b) the mass percentage of the solute c) the molality of the solution d) the mole fraction of the solute

Ans: a) mass solution = 5. 0 g + 225 g = 230. g Vsol. = 230 Ans: a) mass solution = 5.0 g + 225 g = 230. g Vsol. = 230.g x 1 mL = 263 mL 0.876 g moles toluene = (5.0 g C7H8)(1 mol C7H8) = 0.054 mol 92 g C7H8 M = 0.054 mol = 0.21 mol/L .263 L b) moles C6H6 = 225 g x 1mol = 2.88 mol C6H6 80.108 g χC7H6 = 0.054 = .018 0.054 + 2.88 c) molality = .054 mol C7H8 = .24 mol/kg .225 kg C6H6 d) mass % = 5.0 g C7H8 x 100 = 5.0 g x 100 5.0 g C7H8 + 225 g C6H6 230. g = 2.2 %

converting molarity to molality mass solvent mass solute molar moles solute mass mass solution density volume solution M

Practice Problems Zumdahl (8th ed.) p. 532 # 36 p. 531 # 13-15 p. 536 # 98