CHEM L5.1 - Concentration of Solutions

• Solute


  A dissolved substance
• Solvent


  A substance capable of dissolving other substance(s)
• Solution


  A homogenous mixture consisting of a solute and a solvent
• Dissolution


  Process where a solute dissolves in a solvent, forming a solution
• Concentration


  Expresses the amount of a substance present in a mixture
• Percentage by composition


  • Ratio of the mass of solute that is present in a solution, relative to the mass of the solution, as a whole
  • Ratio of the volume of solute that is present in a solution, relative to the volume of the solution, as a whole
• Sample problem #1


  1. Given: Mass of solute= 36.5 g, Mass of solution= 355 g

  2. Required: Mass percent (%m/m)

  3. Equation: %m/m= (mass of solute/mass of solution) x 100

  4. Solution: %m/m= (36.5 g/355 g) x 100

  5. Answer: %m/m= 10.3%
• Sample problem #2


  1. Given: Volume of solute= 90 mL, Volume of solution= 3000 mL

  2. Required: Volume percent (%v/v)

  3. Equation: %v/v= (volume of solute/volume of solution) x 100

  4. Solution: %v/v= (90 mL/3000 mL) x 100

  5. Answer: %v/v= 3%
• ppm


  Parts per million
• ppb


  Parts per billion
• 1 ppm = 103 ppb
• Sample problem #3


  1. Given: Mass of solute= 0.600 g, Mass of solution= 277 g

  2. Required: ppm

  3. Equation: ppm= (mass of solute/mass of solution) x 10^6

  4. Solution: ppm= (0.600 g/277 g) x 10^6

  5. Answer: ppm= 2,166 ppm or 2170 ppm
• Sample problem #4


  1. Given: Mass of solute= 0.48 mg, Mass of solution= 50,000 mg

  2. Required: ppb

  3. Equation: ppb= (mass of solute/mass of solution) x 10^9

  4. Solution: ppb= (0.48 mg/50,000 mg) x 10^9

  5. Answer: ppb= 9,600 ppb
• Mole fraction


  Ratio of the moles of a component to the total moles of all components in a mixture
• Sample problem #5


  1. Given: Moles of solute= 0.100 mol, Mass of solvent= 100.0 g

  2. Required: Mole fraction of solute (Xsolute)

  3. Equation: Xsolute= (moles of solute)/(moles of solution)

  4. Solution: Moles of solvent= 100.0 g / 18 g/mol = 5.6 mol

  Xsolute= 0.100 mol / (5.6 mol + 0.100 mol) = 0.0177
• Molarity (M)


  Solute concentration expressed as moles (mol) of solute in Liters (L) of solution
• Sample problem #6


  1. Given: Moles of solute= 0.225 mol, Volume of solution= 2.80 L

  2. Required: Molarity (M)

  3. Equation: M = moles of solute / L of solution

  4. Solution: M = 0.225 mol / 2.80 L

  5. Answer: M = 0.0804 M
• Molality (m)


  Solute concentration expressed as moles (mol) of solute in kilogram (kg) of solvent
• Sample problem #7


  Given: Mass of solute= 298 g, Mass of solvent= 1.20 kg

  Required: Molality (m)

  Equation: m = moles of solute / kg of solvent

  Solution: Moles of solute = 298 g / 58 g/mol = 5.14 mol

  m = 5.14 mol / 1.20 kg

  Answer: m = 4.28 m
• Calculate the molarity (M) of a solution


  1. Determine moles of solute

  2. Determine volume of solution

  3. Calculate M = moles of solute / volume of solution
• Molarity (M)


  Solute concentration expressed as moles of solute per liter of solution
• Molality (m)


  Solute concentration expressed as moles of solute per kilogram of solvent
• Calculate the molality (m) of a solution


  1. Determine moles of solute

  2. Determine mass of solvent

  3. Calculate m = moles of solute / kg of solvent
• Molarity (M) and Molality (m) are different units of concentration
• Molality is affected by dilution
• Molarity and Molality are units of concentration of solutions
• Molality is not affected by dilution
• Applications of molarity and molality


  • Calculating sugar levels
  • Boiling point elevation
  • Freezing point depression
  • Expressing concentration units