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Percent (%) concentration % (w/v) concentration:

Percent (%) concentration % (w/v) concentration: mass of solute in grams contained by 100 mL solution % (w/w) concentration: mass of solute in grams contained in 100 g of solution % (v/v) concentration: volume of solute in mL in 100 mL solution.

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Percent (%) concentration % (w/v) concentration:

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  1. Percent (%) concentration % (w/v) concentration: mass of solute in grams contained by 100 mL solution % (w/w) concentration: mass of solute in grams contained in 100 g of solution % (v/v) concentration: volume of solute in mL in 100 mL solution

  2. parts per million (ppm) are used to express the concentrations of very dilute solution A solution whose solute concentration is 1 ppm contains 1 g of solute for each million (106) grams of solution or, equivalently, 1 mg of solute per kilogram of solution A 2.5-g sample of groundwater was found to contain 5.4 g of Zn2+. What is the concentration of Zn2+ in parts per million?

  3. Mole Fraction, Molarity, and Molality are concentration expressions based on the number of moles of one or more components of the solution The symbol X is used for mole fraction, with a subscript to indicate the component of interest. The mole fraction of HCl is represented as XHCl. The sum of the mole fractions of all components of a solution must equal 1 A solution of hydrochloric acid contains 36 percent HCl by mass. Calculate the mole fraction of HCl in the solution.

  4. Molarity of a solute in a solution is defined as the number of moles of solute per liters of solution A 2.00L solution of hydrochloric acid contains 36 g of HCl. Calculate the molarity of HCl in the solution. 0.99 mol 2.00 L = 0.495= 0.50 M MHCl = ----------------

  5. The molality of a solution, denoted m, is defined as the number of moles of solute per kilogram of solvent A solution of hydrochloric acid contains 36 percent HCl by mass. Calculate the molality of HCl in the solution. (c) What additional information would you need to calculate the molarity of the solution? Mass of water = 100 gSolu – 36 g HCl = 64 g

  6. (a) Calculate the mole fraction of NaOCl in a commercial bleach solution containing 3.62 mass percent NaOCl in water. =0.00900

  7. (b) What is the molality of a solution made by dissolving 36.5 g of naphthalene, C10H8, in 420 g of toluene, C7H8? (b) 0.678 m =0.28477 / .420 0.678024

  8. Given that the density of a solution of 5.0 g of toluene and 225 g of benzene is 0.876 g/mL, calculate (a) the molarity of the solution; (b) the mass percentage of solute.

  9. Calculate the molality of a solution that contains 25 g of H2SO4 dissolved in 80. g of H2O

  10. Calculate the molality of a 10.0% H3PO4 solution in water

  11. Calculate the molality of a solution that contains 51.2 g of naphthalene, C10H—, in 500. mL of carbon tetrachloride. The density of CCl4 is 1.60 g/mL

  12. If 8.32 grams of methanol, CH3OH, are dissolved in 10.3 grams of water, what is the mole fraction of methanol in the solution?

  13. What is the molarity of 2500. mL of a solution that contains 160. grams of NH4NO3?

  14. 11-1Acid and base Reactions 100. mL of 0.100 M HCl solution and 100. mL of 0.100 M NaOH are mixed. What is the molarity of the salt in the resulting solution? Assume that the volumes are additives. 1. Balance Chemical Equation HCl + NaOH  H2O + NaCl Reaction Ratio: 1 mol 1 mol 1 mol 1 mol 0 10.0 m mol 10.0 m mol Start: 10.0 -10.0 Change: -10.0 After reaction: 0.00 0.00 10.0 For HCl: (100 mL)(0.100mol/L) = 10.0 mmol HCl For NaOH = (100. mL)(0.100mol/L) = 10.0 mmol For NaCl: moles of NaCl at the end of reaction = moles of HCl at beginning = 10.0 m mol 10.0 mmol NaCl (100+100) mL solution MNaCl = ---------------------------- = 0.0500

  15. 11-1Acid and base Reactions 80. mL of 0.100 M HCl solution and 100. mL of 0.100 M NaOH are mixed. What is the molarity of the salt in the resulting solution? Assume that the volumes are additives. 1. Balance Chemical Equation HCl + NaOH  H2O + NaCl Reaction Ratio: 1 mol 1 mol 1 mol 1 mol Start: 8.0 m mol 10.0 m mol 0 8.0 m mol Change: -8.0 -8.0 After reaction: 0.00 2.0 8.0 For HCl: (80 mL)(0.100mol/L) = 8.0 mmol HCl For NaOH = (100. mL)(0.100mol/L) = 10.0 mmol For NaCl: moles of NaCl at the end of reaction = moles of HCl at beginning = 8.0 m mol 8.0 mmol NaCl (100+80) mL sol MNaCl = ---------------------------- = 0.044 2.0 mmol NaOH 180 mL solution = 0.011 MNaOH = ------------------------

  16. 100. mL of 0.100 M HCl solution and 100. mL of 0.80 M NaOH are mixed. What is the molarity of the salt in the resulting solution? Assume that the volumes are additives. 1. Balance Chemical Equation HCl + NaOH  H2O + NaCl Reaction Ratio: 1 mol 1 mol 1 mol 1 mol 10.0 m mol 8.0 m mol 0 Start: Change: -8.0 +8.0 -8.0 After reaction: 2.0 0.0 8.0 For HCl: (100 mL)(0.100mol/L) = 10.0 mmol HCl For NaOH = (100. mL)(0.80mol/L) = 8.0 mmol For NaCl: moles of NaCl at the end of reaction = moles of NaOH at beginning = 8.0 m mol 8.0 mmol NaCl (100+100) mL solution MNaCl = ---------------------------- = 0.040 2.0 mmol NaCl (100+100) mL solution MHCl = ---------------------------- = 0.010

  17. 100. mL of 0.100 M HCl solution and 100. mL of 0.80 M NaOH are mixed. What is the molarity of the salt in the resulting solution? Assume that the volumes are additives. 1. Balance Chemical Equation HCl + NaOH  H2O + NaCl Reaction Ratio: 1 mol 1 mol 1 mol 1 mol 10.0 m mol 80. m mol 0 Start: Change: -10.0 +10.0 -10.0 After reaction: 0.00 70. 8.0 For HCl: (100 mL)(0.100mol/L) = 10.0 mmol HCl For NaOH = (100. mL)(0.80mol/L) = 80. mmol For NaCl: moles of NaCl at the end of reaction = moles of NaOH at beginning = 10. m mol 10. mmol NaCl (100+100) mL solution MNaCl = ---------------------------- = 0.050 70. mmol NaCl (100+100) mL solution MNaOH = ---------------------------- = 0.035

  18. 100. mL of 1.00 M H2SO4 solution is mixed with 200. mL of 1.00 M KOH. What is the molarity of the salt in the resulting solution? Assume that the volumes are additives. 1. Balance Chemical Equation H2SO4 + 2KOH  2H2O + K2SO4 Reaction Ratio: 1 mol 2 mol 2 mol 1 mol Start: 100 m mol 200 m mol 0 100 m mol Change: -200 -100 After reaction: 0.00 0.0 100 Moles of H2SO4: (100 mL)(1.00mol/L) = 100 mmol mols KOH = (200. mL)(1.00mol/L) = 200 mmol moles of K2SO4 at the end of reaction = moles of H2SO4 at beginning = 100 m mol 100 mmol K2SO4 (100+200) mL sol M H2SO4 = --------------------- = 0.333

  19. Titration Titration is a process in which a solution of one reactant, titrant, is carfully added

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