Study Notes: Solutions and Mixtures

Overview of Recruiting Visit and Motivation

  • A recruiting representative provided a campus tour.
  • Morning visits are typically busy, leading to the need for afternoon appointments.
  • Discussion about the impact of early semester visits on motivation levels.

Chapter 8: Solutions and Mixtures

  • Focus on solutions and mixtures; how molecules interact:
    • Solute: The substance that is dissolved (e.g., ethanol, sugar).
    • Solvent: The substance in which the solute is dissolved (often water).
  • Molecules categorized by bonding type:
    • Covalently bonded molecules: Remain bonded in solution (e.g., ethanol, sugar).
    • Ionic compounds: Break into ions.
    • Example: Sodium chloride dissociates into Na⁺ and Cl⁻; acts as an electrolyte.
  • Maximum solubility of solutes discussed.
  • Units of concentration covered:
    • Conversions of concentrations were addressed along with examples.

Sample Calculations

Example 1: Lead Concentration in Groundwater

  • Given: 71.2 g groundwater and 13.85 ppm lead.
  • Calculation goal: Mass of lead in the sample in milligrams.
    1. Formula: ext{PPM} = rac{ ext{grams of solute}}{ ext{grams of solvent}} imes 10^6
    2. Relationship: Since 1 g of water ≈ 1 mL, 71.2 g water correlates to 71.2 mL.
    3. Rearranging the equation yields:
    • ext{Grams of lead} = rac{13.85 ext{ PPM} imes 71.2 ext{ mL}}{10^6}
    1. Result:
    • Grams of lead: 9.86 imes 10^{-4} ext{ g}
    • Convert to milligrams: 9.86 imes 10^{-4} ext{ g} imes 1000 = 0.986 ext{ mg}.

Example 2: Solute in Dilute Solution

  • Problem: Calculate grams of solute in 825 mL of 0.54 M KBr.

    1. Molarity (M): ext{M} = rac{ ext{moles of solute}}{ ext{liters of solution}}.
    2. Find molar mass of KBr ( ext{K} + ext{Br}):
    • Potassium: approximate 39 g.
    • Bromine: approximate 80 g.
    • Total: 119 g/mol.

    1. Convert 825 mL to liters: 0.825 L.

    2. Calculation setup:

      ext{Moles of KBr} = 0.54 ext{ moles/L} imes 0.825 ext{ L}
      = 0.4455 moles.

    3. Calculate grams:

    • 0.4455 ext{ moles} imes 119 ext{ g/mole} = 53.00 ext{ g KBr}.

Example 3: Molarity Calculation from Grams

  • Given: 15.5 g NaCl in 789 mL solution.
  • Find molarity (M).
    1. Convert volume to liters: 0.789 L.
    2. Find molar mass of NaCl: approx 58.44 g/mol.
    3. Calculation Setup:
      ext{Molarity} (M) = rac{15.5 ext{ g NaCl} imes (1 ext{ mol}/58.44 ext{ g})}{0.789 ext{ L}}
      = 0.329 M.

IV Dosage Calculation Examples

  • Example 4: A patient has a blood glucose level of 1.3 g/dL. Convert to mg/mL:
    • Convert g to mg: 1.3 ext{ g} imes 1000 = 1300 ext{ mg}.
    • Convert dL to mL: 1 ext{ dL} = 100 ext{ mL}.
    • Therefore, rac{1300 ext{ mg}}{100 ext{ mL}} = 13 ext{ mg/mL}.

Dosage Calculations in Medicine

  • Oral medications: Example calculation with acetaminophen concentration 21.7 mg/mL, dosage of Nyquil 15 mL.
    1. Calculation: 15 ext{ mL} imes 21.7 ext{ mg/mL} = 325.5 ext{ mg acetaminophen}.

IV Flow Rate Calculations

  • Example: Administration of 25 mg IV over 3 hours; concentration of IV solution is 2.75 mg/mL.
    1. Calculate volume in mL:
    • 25 ext{ mg} imes rac{1 ext{ mL}}{2.75 ext{ mg}} = 9.09 ext{ mL}.
    1. Flow rate:
    • rac{9.09 ext{ mL}}{3 ext{ hours}} = 3.03 ext{ mL/hour}.
  • Another example with flow rate of 10 mL/min; concentration 2.55 mg/mL: Delivery rate = 25.5 mg/min.

Dosage Calculations without Direct Relationships

  • Conversion of symptoms requiring careful analysis of units to arrive at suitable dosage:
    • Example: 625 mg antibiotic ratio to 225 mg/5 mL:
      rac{625 ext{ mg}}{ rac{225 ext{ mg}}{5 ext{ mL}}} o 13.89 ext{ mL}.
    • Express maximum daily dosage in grams per pound.

Dilution Calculations

  • Dilution Concept: Preparing less concentrated solutions from a stock solution.
  • Formula: M1 V1 = M2 V2
    • Use this for solving dilution problems, where $M1$ is the greater concentration (stock) and $M2$ is the lesser concentration.
  • Example: Dilute 0.4 M lead nitrate to a target of 0.2 M:
    • Calculation conclusion shows the relationship of concentration and volume for achieving specific targets.
  • Conclusion: Physics of dilutions aims at achieving a substance concentration conducive for specific applications.

Final Thoughts and Administrative Notes

  • Upcoming chapters to discuss osmosis, dialysis, and transition towards solutions involving weak acids and bases.
  • Important reminders about final exams, class scheduling, and upcoming class topics as the semester progresses.