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Begin by setting up the problem. You are given the mass of ethanol and the volume of

Given: 27.5 g C 2 H 6 O 175 mL H 2 O Find : mass percent Equation and Conversion Factor :. Solution :. EXAMPLE 13.1. Calculating Mass Percent.

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Begin by setting up the problem. You are given the mass of ethanol and the volume of

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  1. Given: 27.5 g C2H6O 175 mL H2O Find:mass percent Equation and Conversion Factor: Solution: EXAMPLE 13.1 Calculating Mass Percent Calculate the mass percent of a solution containing 27.5 g of ethanol (C2H6O) and 175 mL of H2O (Assume that the density of water is 1.00 g/mL.) Begin by setting up the problem. You are given the mass of ethanol and the volume of water and asked to find the mass percent of the solution. You will need the equation that defines mass percent and the density of water. To find the mass percent, substitute into the equation for mass percent. You need the mass of the solution, which is simply the mass of ethanol plus the mass of water. The mass of water is obtained from the volume of water by using the density as a conversion factor. Finally, substitute the correct quantities into the equation and calculate the mass percent

  2. FOR MORE PRACTICE Example 13.11; Problems 43, 44, 45, 46, 47, 48. SKILLBUILDER 13.1 Calculating Mass Percent Calculate the mass percent of a sucrose solution containing 11.3 g of sucrose and 412.1 mL of water. (Assume that the density of water is 1.00 g/mL) EXAMPLE 13.1 Calculating Mass Percent Continued

  3. Given: 11.5% C12H22O11 by mass 85.2 g C12H22O11 Find:mL solution (soft drink) Conversion Factors: Solution Map: EXAMPLE 13.2 Using Mass Percent in Calculations A soft drink contains 11.5% sucrose (C12H22O11) by mass. What volume of the soft drink solution in milliliters contains 85.2 g of sucrose? (Assume a density of 1.00 g/mL). You are given the concentration of sucrose in a soft drink and a mass of sucrose. You are asked to find the volume of the soft drink that contains the given mass of sucrose. Write the mass percent concentration of sucrose as a conversion factor, remembering that percent means per hundred. You will also need the density to use as a conversion factor between mass and volume of the soft drink. Convert from g solute (C12H22O11) to g solution using the mass percent in fractional form as the conversion factor. Convert to mL using the density.

  4. Solution: FOR MORE PRACTICE Example 13.12; Problems 49, 50, 51, 52, 53, 54. SKILLBUILDER 13.2 Using Mass Percent in Calculations How much sucrose (C12H22O11) in grams is contained in 355 mL (12 oz) of the soft drink in Example 13.2? EXAMPLE 13.2 Using Mass Percent in Calculations Continued Follow the solution map to solve the problem.

  5. Given: 15.5 g NaCl 1.50 L solution Find:molarity (M) Conversion Factors: Solution: EXAMPLE 13.3 Calculating Molarity Calculate the molarity of a solution made by putting 15.5 g NaCl into a beaker and adding water to make 1.50 L of NaCl solution. You are given the mass of sodium chloride (the solute) and the volume of solution. You are asked to find the molarity of the solution. You will need the equation that defines molarity and the molar mass of NaCl. To calculate molarity, substitute the correct values into the equation and compute the answer. However, you must first convert the amount of NaCl from grams to moles using the molar mass of NaCl.

  6. FOR MORE PRACTICE Example 13.13; Problems 61, 62, 63, 64, 65, 66. SKILLBUILDER 13.3 Calculating Molarity Calculate the molarity of a solution made by putting 55.8 g of NaNO3 into a beaker and diluting to 2.50 L. EXAMPLE 13.3 Calculating Molarity Continued

  7. Given: 0.114 M NaOH 1.24 mol NaOH Find:L solution Conversion Factor: Solution Map: Solution: EXAMPLE 13.4 Using Molarity in Calculations How many liters of a 0.114 M NaOH solution contains 1.24 mol of NaOH? You are given the molarity of an NaOH solution and the number of moles of NaOH. You are asked to find the volume of solution that contains the given number of moles. You will need to use the molarity of the solution (that is given) as a conversion factor The solution map begins with mol NaOH and shows the conversion to liters of solution using the molarity. Solve the problem by following the solution map.

  8. FOR MORE PRACTICE Example 13.14; Problems 67, 68, 69, 70, 71, 72. SKILLBUILDER 13.4 Using Molarity in Calculations How much of a 0.225 M KCl solution contains 55.8 g of KCl? EXAMPLE 13.4 Calculating Molarity Continued

  9. FOR MORE PRACTICE Problems 79, 80, 81, 82. SKILLBUILDER 13.5 Ion Concentration Determine the molar concentrations of Ca2+ and Cl– in a 0.75 M CaCl2 solution. EXAMPLE 13.5 Ion Concentration Determine the molar concentrations of Na+ and PO43– in a 1.50 M Na3PO4 solution. You are given the concentration of an ionic solution and asked to find the concentrations of the component ions. Given: 1.50 M Na3PO4 Find:molarity (M) of Na+ and PO43– Since a formula unit of Na3PO4 contains 3 Na+ ions (as indicated by the subscript), the concentration of Na+ is three times the concentration of Na3PO4. Since the same formula unit contains one PO43– ion, the concentration of PO43– is equal to the concentration of Na3PO4. Solution: molarity of Na+ = 3(1.50 M) = 4.50 M molarity of PO43– = 1.50 M

  10. FOR MORE PRACTICE Example 13.15; Problems 83, 84, 85, 86, 87, 88, 89, 90. Solution: SKILLBUILDER 13.6 Solution Dilution How much 6.0 M NaNO3 solution should be used to make 0.585 L of a 1.2 M NaNO3 solution? EXAMPLE 13.6 Solution Dilution To what volume should you dilute 0.100 L of a 15 M NaOH solution to obtain a 1.0 M NaOH solution? You are given the the initial volume and concentration of an NaOH solution and a final concentration. You are asked to find the volume required to dilute the solution to the given final concentration. The necessary equation is the solution dilution equation given previously in this section. Given: V1 = 0.100 L M1 = 15 M M2 = 1.0 M Find:V2 Equation: M1V1 = M2V2 Solve the equation for V2 the volume of the final solution, and substitute the required quantities to compute V2. You would make the solution by diluting 0.100 L of the stock solution to a total volume of 1.5 L(V2). The resulting solution will have a concentration of 1.0 M.

  11. Consider the following precipitation reaction: How much 0.115 M KI solution in liters is required to completely precipitate the Pb2+ in 0.104 L of 0.225 M Pb(NO3)2 solution? Given: 0.115 M KI 0.104 L Pb(NO3)2 solution 0.225 M Pb(NO3)2 Find:L KI solution Conversion Factors: EXAMPLE 13.7 Solution Stoichiometry You are given the concentration of a reactant, KI, in a chemical reaction. You are also given the volume and concentration of a second reactant, Pb(NO3)2 You are asked to find the volume of the first reactant that completely reacts with the given amount of the second. The conversion factors for this problem are the molarities of the two solutions expressed in mol of solute per L of solution and the stoichiometric relationship (from the balanced equation) between mol KI and mol Pb(NO3)2.

  12. Solution Map: Solution: SKILLBUILDER 13.7 Solution Stoichiometry How many milliliters of 0.112 M Na2CO3 are necessary to completely react with 27.2 mL of 0.135 M HNO3 according to the following reaction? EXAMPLE 13.7 Solution Stoichiometry Continued The solution map for this problem is similar to the solution maps for other stoichiometric problems. First use the volume and molarity of Pb(NO3)2 solution to get mol Pb(NO3)2 Then use the stoichiometric coefficients from the equation to convert mol Pb(NO3)2 to mol KI. Finally, use mol KI to find L KI solution Follow the solution map to solve the problem. Begin with volume of Pb(NO3)2 solution and cancel units to arrive at volume of KI solution

  13. A 25.0-mL sample of HNO3 solution requires 35.7 mL of 0.108 M Na2CO3 to completely react with all of the HNO3 in the solution. What was the concentration of the HNO3 solution? SKILLBUILDER PLUS FOR MORE PRACTICE Example 13.16; Problems 91, 92, 93, 94. EXAMPLE 13.7 Solution Stoichiometry Continued

  14. Given: 17.2 g C2H6O2 0.500 kg H2O Find:molality (m) Equation and Conversion Factor: FOR MORE PRACTICE Example 13.17; Problems 99, 100, 101, 102. Solution: SKILLBUILDER 13.8 Calculating Molality Calculate the molality (m) of a sucrose (C12H22O11) solution containing 50.4 g sucrose and 0.332 kg of water. EXAMPLE 13.8 Calculating Molality Calculate the molality of a solution containing 17.2 g of ethylene glycol (C2H6O2) dissolved in 0.500 kg of water. You are given the mass of ethylene glycol in grams and the mass of the solvent in kilograms. You are asked to find the molality of the resulting solution. You will need the equation that defines molality and the molar mass of ethylene glycol. To calculate molality, simply substitute the correct values into the equation and compute the answer. However, you must first convert the amount of C2H6O2 from grams to moles using the molar mass of C2H6O2.

  15. FOR MORE PRACTICE Example 13.18; Problems 103, 104. Solution: Freezing point = 0.00 °C – 3.2 °C = –3.2 ° SKILLBUILDER 13.9 Freezing Point Depression Calculate the freezing point of a 2.6 m sucrose solution. EXAMPLE 13.9 Freezing Point Depression Calculate the freezing point of a 1.7 m ethylene glycol solution. You are given the molality of an aqueous solution and asked to find the freezing point depression. You will need the freezing point depression equation given previously in this section. Given: 1.7 m solution Find:ΔTf Equation: ΔTf = m × Kf To solve this problem, simply substitute the values into the equation for freezing point depression and calculate ΔTf The actual freezing point will be the freezing point of pure water (0.00 °C) – ΔTf

  16. FOR MORE PRACTICE Problems 105, 106, 107, 108. Solution: SKILLBUILDER 13.10 Boiling Point Elevation Calculate the boiling point of a 3.5 m glucose solution. EXAMPLE 13.10 Boiling Point Elevation Calculate the boiling point of a 1.7 m ethylene glycol solution. You are given the molality of an aqueous solution and asked to find the boiling point depression. You will need the boiling point elevation equation given previously in this section. Given: 1.7 m solution Find:boiling point Equation: ΔTb= m × Kb To solve this problem, simply substitute the values into the equation for boiling point elevation and calculate ΔTb The actual boiling point of the solution will be the boiling point of pure water (100.00 °C) plus ΔTb

  17. Given: 19 g solute 158 g solvent Find: mass percent Equation: Solution: EXAMPLE 13.11 Calculating Mass Percent Find the mass percent concentration of a solution containing 19 g of solute and 158 g of solvent.

  18. Given: 5.80% KCl by mass 0.337 L solution Find: g KCl Conversion Factors: Solution Map: Solution: EXAMPLE 13.12 Using Mass Percent in Calculations How much KCl in grams is in 0.337 L of a 5.80% mass percent KCl solution? (Assume that the density of the solution is 1.05 g/mL.)

  19. Given: 0.22 mol KCl 0.455 L solution Find: molarity (M) Equation: Solution: EXAMPLE 13.13 Calculating Molarity Calculate the molarity of a KCl solution containing 0.22 mol of KCl in 0.455 L of solution.

  20. Given: 1.25 M KCl 0.488 L solution Find: g KCl Conversion Factors: Solution Map: Solution: EXAMPLE 13.14 Using Molarity in Calculations How much KCl in grams is contained in 0.488 L of 1.25 M KCl solution?

  21. Given: M1 = 8.0 M M2 = 2.7 M V2 = 0.400 L Find: V1 Equation: M1V1 = M2V2 Solution: EXAMPLE 13.15 Solution Dilution How much of an 8.0 M HCl solution should be used to make 0.400 L of a 2.7 M HCl solution?

  22. Consider the following reaction: How much 0.113 M NaOH solution is required to completely neutralize 1.25 L of 0.228 M HCl solution? Given: 1.25 L HCl solution 0.228 M HCl 0.113 M NaOH Find: L NaOH solution Conversion Factors: Solution Map: EXAMPLE 13.16 Solution Stoichiometry

  23. Solution: EXAMPLE 13.16 Solution Stoichiometry Continued

  24. Given: 0.183 mol of sucrose 1.10 kg H2O Find: molality (m) Equation: Solution: EXAMPLE 13.17 Calculating Molality Calculate the molality of a solution containing 0.183 mol of sucrose dissolved in 1.10 kg of water.

  25. Given: 2.5 m solution Find: ΔTf Equation: ΔTf = m × Kf Solution: Freezing Point = 0.00 °C – 4.7 °C = –4.7 °C EXAMPLE 13.18 Freezing Point Depression and Boiling Point Elevation Calculate the freezing point of a 2.5 m aqueous sucrose solution.

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