Comprehensive Study Guide to Solutions and Solution Chemistry

Principles of Solution Formation and Intermolecular Forces

• Fundamentals of Solution Formation:

  • A solution forms when the attractive forces between solute and solvent particles are sufficiently large to overcome both solute–solute interactions and solvent–solvent interactions.
  • Mutual solubility is governed by the rule of "like dissolves like", meaning that substances with similar types of intermolecular forces will dissolve in one another.

• Solubility Compatibility Matrix:

  • Polar solute + Polar solvent: Solution forms.
  • Nonpolar solute + Nonpolar solvent: Solution forms.
  • Polar solute + Nonpolar solvent: Solution will not form.
  • Nonpolar solute + Polar solvent: Solution will not form.

• Hierarchy of Chemical and Intermolecular Attractive Forces:
  • Forces Between Atoms or Ions (Strong):
  • Ionic Bond: Electrostatic attraction between full positive and negative ions arranged in an alternating crystal lattice (e.g., $\text{Na}^+\text{Cl}^-$).
  • Covalent Bond: Shared pairs of electrons between nonmetal atoms ($X:X$, e.g., $\text{Cl}-\text{Cl}$).
  • Forces Between Molecules (Weaker):
  • Hydrogen Bond: A strong dipole attraction occurring when hydrogen is covalently bonded to highly electronegative atoms ($\text{F}$, $\text{O}$, or $\text{N}$), represented as $\text{H}^{\delta+}-\text{X}^{\delta-} \cdots \text{H}^{\delta+}-\text{X}^{\delta-}$ (e.g., $\text{H}^{\delta+}-\text{F}^{\delta-} \cdots \text{H}^{\delta+}-\text{F}^{\delta-}$).
  • Dipole–Dipole Attractions: Electrostatic attractions between permanent dipoles of polar molecules where $X$ and $Y$ are nonmetals with different electronegativities, represented as $\text{Y}^{\delta+}\text{X}^{\delta-} \cdots \text{Y}^{\delta+}\text{X}^{\delta-}$ (e.g., $\text{Br}^{\delta+}-\text{Cl}^{\delta-} \cdots \text{Br}^{\delta+}-\text{Cl}^{\delta-}$).
  • Dispersion Forces: Weak attractions resulting from temporary shifts of electron density causing temporary dipoles in nonpolar molecules, represented as $\text{X}:\text{X} \cdots \text{X}:\text{X}$ (e.g., $\text{F}^{\delta+}-\text{F}^{\delta-} \cdots \text{F}^{\delta+}-\text{F}^{\delta-}$).

Practical Applications of "Like Dissolves Like"

• Waterproof Cosmetics:
  • Waterproof mascara resists smearing from tears, rain, or swimming because it consists of nonpolar substances.
  • Formulated with waxes and oils containing exclusively nonpolar $\text{C}-\text{H}$ bonds.
  • Specific nonpolar components include:
  • Petroleum distillates: Hydrocarbon mixtures.
  • Polyethylene: Synthetic polymers composed of continuous carbon chains with two hydrogen atoms bonded to each carbon atom.
  • Removal requires nonpolar, oil-based cleansing agents rather than water.

• Permanent Markers:

  • Permanent marker ink resists removal by water because water is highly polar while ink components are nonpolar.
  • Key ingredients in permanent marker ink include:
  • Glyceride
  • Pyrrolidone
  • Resin
  • Colorant
  • Removal Mechanism: Rubbing alcohol (isopropanol/ethanol) possesses both polar and nonpolar characteristics, making it less polar than water. It can interact with nonpolar ink molecules to dissolve and lift the pigment from surfaces.

• Capsaicin and Superhydrophobic Properties:

  • Capsaicin (the spicy component in hot peppers) is predominantly nonpolar and insoluble in water; milk or fatty substances are required to alleviate burning sensations.
  • Superhydrophobic materials utilize extreme nonpolar surface textures to repel water entirely, causing water droplets to bead and roll off without wetting.

Dissolution of Ionic Compounds and Hydration

• Hydration Process of Ionic Salts:
  • When an ionic solid such as sodium chloride ($\text{NaCl}$) is added to water, water molecules orient themselves around each ion on the surface of the lattice.
  • Partially negative oxygen atoms of water attract positive sodium ions ($\text{Na}^+$).
  • Partially positive hydrogen atoms of water attract negative chloride ions ($\text{Cl}^-$).
  • Water molecules pull ions away from the crystal lattice into solution, forming hydrated ions surrounded by a hydration shell:

$\text{NaCl}(s) \rightarrow \text{Na}^+(aq) + \text{Cl}^-(aq)$

Factors Affecting Solubility: Temperature and Pressure

• Definition of Solubility:
  • Solubility is defined as the maximum amount of solute that dissolves in a specific amount of solvent at a given temperature.
  • Commonly expressed as grams of solute per $100\,\text{g}$ of solvent (typically water):

$\frac{\text{g of solute}}{100\,\text{g water}}$

• Effect of Temperature:

  • Solids: The solubility of most solids increases as temperature increases.
  • Gases: The solubility of gases decreases as temperature increases. Heating a gas–liquid solution imparts kinetic energy to the gas molecules, allowing them to escape the liquid phase into the gas phase.

• Effect of Pressure and Henry's Law:

  • Pressure has a negligible effect on the solubility of solids and liquids, but profoundly impacts gases.
  • Henry's Law: The solubility of a gas in a liquid is directly proportional to the partial pressure of that gas above the liquid.
  • Carbonated Beverage System:
  • In a sealed can or bottle, $\text{CO}_2$ gas is kept under high pressure above the liquid, forcing a high concentration of $\text{CO}_2$ molecules to dissolve.
  • When the container is opened, the pressure drops to atmospheric pressure, decreasing $\text{CO}_2$ solubility and causing gas molecules to bubble out of solution.

Supersaturation and Candy Chemistry

• Classification of Confections:
  • Crystalline Candies:
  • Examples: Fudge, nougat, fondant.
  • Contain many fine, small sucrose crystals in a smooth, creamy texture.
  • Produced from solutions boiled at lower temperatures with lower sugar concentrations.
  • Formed by slow cooling without stirring to permit orderly sucrose lattice formation.
  • Non-Crystalline (Amorphous) Candies:
  • Examples: Lollipops, candy canes, caramel.
  • Hard and brittle texture containing no crystals.
  • Produced from solutions boiled at higher temperatures with higher sugar concentrations.
  • Formed from very saturated solutions where crystallization is inhibited by rapid cooling, high viscosity, or the introduction of interfering agents (such as glucose and fructose).

• Rock Candy (Supersaturated Solutions):

  • Prepared by dissolving sugar in water at elevated temperatures to produce a saturated solution, followed by undisturbed cooling.
  • As temperature falls, the solution holds more dissolved solute than normal saturation limits allow, entering an unstable supersaturated state.
  • Excess solute precipitates onto seed points, growing large sugar crystals outside the solution until stability is restored.

• Pop Rocks:

  • Manufactured by heating a sucrose solution and exposing it to carbon dioxide gas under high pressure (nearly $40\,\text{times}$ atmospheric pressure).
  • Rapid cooling traps high-pressure $\text{CO}_2$ bubbles inside sugar crystals.
  • When consumed, mechanical chewing or salivary amylase dissolves the sugar matrix, releasing the trapped $\text{CO}_2$ bubbles with an explosive popping sound.

Clinical Significance of Solubility: Uric Acid and Gout

• Pathophysiology of Gout:
  • Gout is a metabolic disorder primarily affecting adult men over $40\,\text{years}$ of age.
  • Occurs when uric acid concentration in blood plasma exceeds its solubility threshold of $7\,\text{mg} / 100\,\text{mL}$ of plasma at body temperature ($37\,^\circ\text{C}$).
  • When solubility is exceeded, needle-like uric acid crystals precipitate out of solution and deposit into joint tissues, particularly the big toe and fingers, causing severe inflammation and pain.

Units of Solution Concentration and Calculations

• General Concentration Definition:

$\text{Concentration} = \frac{\text{amount of solute}}{\text{amount of solution}}$

• Mass Percent (% m/m):
  • Measures the grams of solute present in $100\,\text{g}$ of solution:

$\text{mass percent (m/m)} = \frac{\text{g of solute}}{\text{g of solute} + \text{g of solvent}} \times 100\%$\text{mass percent (m/m)} = \frac{\text{g of solute}}{100\,\text{g of solution}}\n\n* **Volume Percent (% v/v):**\n * Measures the milliliters of liquid solute in 100\,\text{mL} of solution:\n\n\text{volume percent (v/v)} = \frac{\text{mL of solute}}{\text{mL of solution}} \times 100\%$\text{volume percent (v/v)} = \frac{\text{mL of solute}}{100\,\text{mL of solution}}$

• Mass/Volume Percent (% m/v):
  • Measures the grams of solute dissolved in $100\,\text{mL}$ of solution:

$\text{mass/volume percent (m/v)} = \frac{\text{g of solute}}{\text{mL of solution}} \times 100\%$\text{mass/volume percent (m/v)} = \frac{\text{g of solute}}{100\,\text{mL of solution}}\n\n* **Molarity (M):**\n * Defined as moles of solute per liter of total solution:\n\nM = \frac{\text{moles of solute}}{\text{liter of solution}}$1.0\,M\,\text{NaCl} = \frac{1.0\,\text{mole NaCl}}{1\,\text{L solution}}$

• Parts per Million (ppm) and Parts per Billion (ppb):
  • Utilized for very dilute solutions:

$\text{ppm} = \frac{\text{mass of solute (g)}}{\text{mass of solution (g)}} \times 10^6$\text{ppm} = \frac{\text{volume of solute (mL)}}{\text{volume of solution (mL)}} \times 10^6\n\n * For parts per billion (ppb), the ratio is multiplied by 10^9$($1{,}000{,}000{,}000).\n\n# Electrolytes, Equivalents, and Physiological Fluid Balance\n\n* **Equivalents (Eq) and Milliequivalents (mEq):**\n * Equivalents express the molar concentration of ionic charge in solution:\n * 1\,\text{M } \text{Na}^+\text{ solution} = 1\,\text{Eq/L } \text{Na}^+\n * 1\,\text{M } \text{Ca}^{2+}\text{ solution} = 2\,\text{Eq/L } \text{Ca}^{2+}\n * 1\,\text{Eq} = 1000\,\text{mEq}\n * Intravenous and physiological electrolyte concentrations are typically reported in \text{mEq/L}.\n\n* **Normal Serum Electrolyte Concentration Ranges:**\n * **Sodium (\text{Na}^+$):**$135\text{--}147\,\text{mEq/L}\n * **Chloride (\text{Cl}^-$):**$98\text{--}106\,\text{mEq/L}\n * **Potassium (\text{K}^+$):**$3.5\text{--}5.5\,\text{mEq/L}\n * **Calcium (\text{Ca}^{2+}$):**$4.5\text{--}5.5\,\text{mEq/L}\n * **Magnesium (\text{Mg}^{2+}$):**$1.5\text{--}3.0\,\text{mEq/L}\n\n* **Clinical Terminology for Electrolyte Imbalances:**\n * **Calcium:** Hypercalcemia (high) / Hypocalcemia (low)\n * **Chloride:** Hyperchloremia (high) / Hypochloremia (low)\n * **Magnesium:** Hypermagnesemia (high) / Hypomagnesemia (low)\n * **Phosphate:** Hyperphosphatemia (high) / Hypophosphatemia (low)\n * **Potassium:** Hyperkalemia (high) / Hypokalemia (low)\n * **Sodium:** Hypernatremia (high) / Hyponatremia (low)\n\n* **Composition of Intravenous Solutions:**\n * **Sodium chloride (0.9%):** \text{Na}^+\,154\,\text{mEq/L}$,$\text{Cl}^-\,154\,\text{mEq/L}. Used for replacement of fluid loss.\n * **Potassium chloride with 5.0% dextrose:** \text{K}^+\,40\,\text{mEq/L}$,$\text{Cl}^-\,40\,\text{mEq/L}. Used for treatment of malnutrition (hypokalemia).\n * **Ringer's solution:** \text{Na}^+\,147\,\text{mEq/L}$,$\text{K}^+\,4\,\text{mEq/L}$,$\text{Ca}^{2+}\,4\,\text{mEq/L}$,$\text{Cl}^-\,155\,\text{mEq/L}. Used for replacement of fluids and electrolytes lost through dehydration.\n * **Maintenance solution with 5.0% dextrose:** \text{Na}^+\,40\,\text{mEq/L}$,$\text{K}^+\,35\,\text{mEq/L}$,$\text{Cl}^-\,40\,\text{mEq/L}$,$\text{lactate}^-\,20\,\text{mEq/L}$,$\text{HPO}4^{2-}\,15\,\text{mEq/L}. Used for maintenance of fluid and electrolyte levels.\n * **Replacement solution (extracellular):** \text{Na}^+\,140\,\text{mEq/L}$,$\text{K}^+\,10\,\text{mEq/L}$,$\text{Ca}^{2+}\,5\,\text{mEq/L}$,$\text{Mg}^{2+}\,3\,\text{mEq/L}$,$\text{Cl}^-\,103\,\text{mEq/L}$,$\text{acetate}^-\,47\,\text{mEq/L}$,$\text{citrate}^{3-}\,8\,\text{mEq/L}. Used for replacement of electrolytes in extracellular fluids.\n\n\n\n* **Electrolyte Balance in Human Blood Plasma:**\n * Blood plasma maintains electrical neutrality: total positive charges equal total negative charges.\n * **Cations Total:** 150\,\text{mEq/L}\n * \text{Na}^+: 138\,\text{mEq/L}\n * \text{K}^+: 5\,\text{mEq/L}\n * \text{Ca}^{2+}: 4\,\text{mEq/L}\n * \text{Mg}^{2+}: 3\,\text{mEq/L}\n * **Anions Total:** 150\,\text{mEq/L}\n * \text{Cl}^-: 110\,\text{mEq/L}\n * \text{HCO}_3^-: 30\,\text{mEq/L}\n * \text{Proteins}: 6\,\text{mEq/L}\n * \text{HPO}_4^{2-}: 4\,\text{mEq/L}\n\n\n\n# Solution Dilution Principles and Calculations\n\n* **Dilution Process:**\n * When additional solvent (water) is added to a solution:\n * Solution volume increases.\n * Solute concentration decreases.\n * The total mass and total moles of solute remain constant.\n\n* **Dilution Mathematical Equations:**\n * For percent concentrations:\n\nC_1V_1 = C_2V_2\n\n * For molarity:\n\nM_1V_1 = M_2V_2\n\n# Colligative Properties and Osmotic Phenomena\n\n* **Definition of Colligative Properties:**\n * Solution properties that depend solely on the total concentration (number) of solute particles present, rather than their chemical identity.\n * **Four Primary Colligative Properties:**\n * Vapor pressure lowering\n * Freezing point depression\n * Boiling point elevation\n * Osmotic pressure\n\n* **Osmosis Mechanism:**\n * Water (solvent) molecules diffuse across a semipermeable membrane from a region of lower solute concentration (higher water concentration) to higher solute concentration (lower water concentration).\n * Over time, the liquid level of the more concentrated solution rises until hydrostatic pressure counters the osmotic driving force.\n\n\n\n# Tonicity, Osmotic Pressure, and Cellular Effects\n\n* **Isotonic Solutions:**\n * Exert the identical osmotic pressure as intracellular body fluids.\n * Standard clinical isotonic intravenous solutions:\n * 0.90\%\,(\text{m/v})\,\text{NaCl} (normal saline)\n * 5.0\%\,(\text{m/v})\,\text{glucose} (dextrose)\n * Red blood cells placed in an isotonic environment maintain constant shape and volume due to equal rates of water influx and efflux.\n\n\n\n* **Hypotonic Solutions:**\n * Possess a lower solute concentration (lower osmotic pressure) than red blood cell contents.\n * Water diffuses into the cell by osmosis.\n * The influx of water causes cells to swell and burst, a phenomenon known as **hemolysis**.\n\n\n\n* **Hypertonic Solutions:**\n * Possess a higher solute concentration (higher osmotic pressure) than red blood cell contents.\n * Water diffuses out of the cell into the hypertonic solution by osmosis.\n * Loss of water causes cells to shrivel and shrink, a phenomenon known as **crenation**.\n\n\n\n* **Physiological Hazard of Ingesting Seawater:**\n * Ocean seawater has an osmolarity roughly three times greater than human bodily fluids.\n * Ingesting seawater creates a severely hypertonic environment in the gastrointestinal tract, causing water to leave cells via osmosis to dilute the high salt content, worsening cellular dehydration and systemic thirst.\n\n# Practice Exercises and Clicker Questions\n\n* **Water Solubility Screening:**\n * *Question:* Which of the following solutes will dissolve in water? Why? a. \text{Na}_2\text{O}$, b.$\text{C}_8\text{H}{18}$, c.$\text{I}2$, d.$\text{HCl}\n * *Analysis:* Water is polar. \text{Na}_2\text{O}$is an ionic compound and$\text{HCl}$is a polar molecular compound that ionizes; both dissolve in water. In contrast, octane ($\text{C}_8\text{H}{18}$) and iodine ($\text{I}_2) are nonpolar molecules with dispersion forces that do not dissolve in polar water.\n\n* **Solubility Calculations:**\n * *Question 1:* If the solubility of sodium nitrate is 114\,\text{g} / 100\,\text{g } \text{H}_2\text{O}$at$50\,^\circ\text{C}$, how many grams of sodium nitrate will dissolve in$275\,\text{g}$of water at$50\,^\circ\text{C}$? (Options: a.$114\,\text{g}$, b.$228\,\text{g}$, c.$314\,\text{g}$, d.$482\,\text{g})\n * *Calculation:*\n\n275\,\text{g } \text{H}_2\text{O} \times \frac{114\,\text{g } \text{NaNO}_3}{100\,\text{g } \text{H}_2\text{O}} = 313.5\,\text{g } \text{NaNO}_3 \approx 314\,\text{g}\n\n * *Answer:* **c. 314\,\text{g}**\n * *Question 2:* If the solubility of silver nitrate (\text{AgNO}_3$) is$122\,\text{g} / 100\,\text{mL}$, how much could be dissolved in$556\,\text{mL}$? (Options: a.$219\,\text{g}$, b.$388\,\text{g}$, c.$456\,\text{g}$, d.$678\,\text{g})\n * *Calculation:*\n\n556\,\text{mL} \times \frac{122\,\text{g } \text{AgNO}_3}{100\,\text{mL}} = 678.32\,\text{g } \text{AgNO}_3 \approx 678\,\text{g}\n\n * *Answer:* **d. 678\,\text{g}**\n\n* **Mass Percent Calculations:**\n * *Question 1:* A solution is prepared by mixing 15.0\,\text{g}$of$\text{Na}_2\text{CO}_3$and$235\,\text{g}$of$\text{H}_2\text{O}$. Calculate the mass percent ($15.0\%\,(\text{m/m})$, b.$6.38\%\,(\text{m/m})$, c.$6.00\%\,(\text{m/m}))\n * *Calculation:*\n\n\text{Total mass of solution} = 15.0\,\text{g} + 235\,\text{g} = 250.0\,\text{g}$\text{Mass percent} = \frac{15.0\,\text{g}}{250.0\,\text{g}} \times 100\% = 6.00\%$

* *Answer:* **c. $6.00\%\,(\text{m/m})\,\text{Na}_2\text{CO}_3$**

• Question 2: How many grams of $\text{KCl}$ are needed to prepare $150.\,\text{g}$ of a $7.25\%\,(\text{m/m})\,\text{KCl}$ solution? (Options: a. $10.9\,\text{g}$, b. $4.83\,\text{g}$, c. $13.8\,\text{g}$, d. $20.7\,\text{g}$)
  • Calculation:

$150.\,\text{g solution} \times \frac{7.25\,\text{g KCl}}{100\,\text{g solution}} = 10.875\,\text{g KCl} \approx 10.9\,\text{g}$

* *Answer:* **a. $10.9\,\text{g}$**

• Question 3: How many grams of $\text{NaOH}$ are needed to prepare $75.0\,\text{g}$ of $14.0\%\,(\text{m/m})\,\text{NaOH}$ solution? (Options: a. $10.5\,\text{g}$, b. $75.0\,\text{g}$, c. $536\,\text{g}$)
  • Calculation:

$75.0\,\text{g solution} \times \frac{14.0\,\text{g NaOH}}{100\,\text{g solution}} = 10.5\,\text{g NaOH}$

* *Answer:* **a. $10.5\,\text{g of NaOH}$**

• Volume Percent Calculations:
  • Question 1: A mouthwash contains $22.5\%\,(\text{v/v})$ alcohol. If the bottle of mouthwash contains $355\,\text{mL}$, what would be the volume, in milliliters, of the alcohol? (Options: a. $46.6\,\text{mL}$, b. $57.7\,\text{mL}$, c. $68.8\,\text{mL}$, d. $79.9\,\text{mL}$)
  • Calculation:

$355\,\text{mL solution} \times \frac{22.5\,\text{mL alcohol}}{100\,\text{mL solution}} = 79.875\,\text{mL} \approx 79.9\,\text{mL}$

* *Answer:* **d. $79.9\,\text{mL of alcohol}$**

• Question 2: How many milliliters of a $5.75\%\,(\text{v/v})$ ethanol solution can be prepared from $2.25\,\text{mL}$ of ethanol? (Options: a. $2.56\,\text{mL}$, b. $12.9\,\text{mL}$, c. $39.1\,\text{mL}$)
  • Calculation:

$2.25\,\text{mL ethanol} \times \frac{100\,\text{mL solution}}{5.75\,\text{mL ethanol}} = 39.13\,\text{mL} \approx 39.1\,\text{mL}$

* *Answer:* **c. $39.1\,\text{mL}$**

• Mass/Volume Percent Calculations:
  • Question 1: A nurse needs to prepare an IV that contains $0.30\%\,(\text{m/v})$ potassium chloride. How much solution can he make if he has $1.2\,\text{g}$ of $\text{KCl}$? (Options: a. $0.036\,\text{mL}$, b. $3.6\,\text{mL}$, c. $4.0\,\text{mL}$, d. $4.0 \times 10^2\,\text{mL}$)
  • Calculation:

$1.2\,\text{g KCl} \times \frac{100\,\text{mL solution}}{0.30\,\text{g KCl}} = 400\,\text{mL} = 4.0 \times 10^2\,\text{mL}$

* *Answer:* **d. $4.0 \times 10^2\,\text{mL}$**

• Question 2: A patient needs $100.0\,\text{g}$ of glucose in the next $12\,\text{hours}$. The number of liters of a $5.0\%\,(\text{m/v})$ glucose solution that must be given is which of the following? (Options: a. $0.50\,\text{L}$, b. $1.0\,\text{L}$, c. $1.5\,\text{L}$, d. $2.0\,\text{L}$)
  • Calculation:

$100.0\,\text{g glucose} \times \frac{100\,\text{mL solution}}{5.0\,\text{g glucose}} = 2000\,\text{mL} = 2.0\,\text{L}$

* *Answer:* **d. $2.0\,\text{L}$**

• Molarity Calculations:
  • Question 1: What is the molarity of $0.500\,\text{L}$ of $\text{NaOH}$ solution if it contains $6.00\,\text{g}$ of $\text{NaOH}$?
  • Calculation:

$\text{Molar mass of NaOH} = 22.99 + 16.00 + 1.01 = 40.00\,\text{g/mol}$\text{Moles of NaOH} = \frac{6.00\,\text{g}}{40.00\,\text{g/mol}} = 0.150\,\text{mol}$M = \frac{0.150\,\text{mol}}{0.500\,\text{L}} = 0.300\,\text{M}$

* *Answer:* **$0.300\,\text{M}$**

• Question 2: What is the molarity of $0.225\,\text{L}$ of a $\text{KNO}_3$ solution containing $34.8\,\text{g}$ of $\text{KNO}_3$? (Options: a. $0.344\,\text{M}$, b. $1.53\,\text{M}$, c. $15.5\,\text{M}$)
  • Calculation:

$\text{Molar mass of KNO}_3 = 39.10 + 14.01 + 3(16.00) = 101.11\,\text{g/mol}$\text{Moles of KNO}_3 = \frac{34.8\,\text{g}}{101.11\,\text{g/mol}} = 0.3442\,\text{mol}$M = \frac{0.3442\,\text{mol}}{0.225\,\text{L}} = 1.5297\,\text{M} \approx 1.53\,\text{M}$

* *Answer:* **b. $1.53\,\text{M}$**

• Question 3: Calculate the molarity of $5.8\,\text{L}$ of $\text{NaCl}$ solution that contains $65\,\text{g}$ of $\text{NaCl}$. (Options: a. $1.1\,\text{M}$, b. $0.19\,\text{M}$, c. $5.3\,\text{M}$, d. $6.5\,\text{M}$)
  • Calculation:

$\text{Molar mass of NaCl} = 22.99 + 35.45 = 58.44\,\text{g/mol}$\text{Moles of NaCl} = \frac{65\,\text{g}}{58.44\,\text{g/mol}} = 1.112\,\text{mol}$M = \frac{1.112\,\text{mol}}{5.8\,\text{L}} = 0.1918\,\text{M} \approx 0.19\,\text{M}$

* *Answer:* **b. $0.19\,\text{M}$**

• Question 4: What is the molarity of $2.0\,\text{moles}$ of glucose in $4.0\,\text{L}$ of glucose solution? (Options: a. $0.25\,\text{M}$, b. $0.50\,\text{M}$, c. $0.75\,\text{M}$, d. $1.00\,\text{M}$)
  • Calculation:

$M = \frac{2.0\,\text{moles}}{4.0\,\text{L}} = 0.50\,\text{M}$

* *Answer:* **b. $0.50\,\text{M glucose}$**

• Question 5: How many grams of $\text{AlCl}_3$ are needed to prepare $125\,\text{mL}$ of a $0.150\,\text{M}$ solution? (Options: a. $20.0\,\text{g}$, b. $16.7\,\text{g}$, c. $2.50\,\text{g}$)
  • Calculation:

$\text{Molar mass of AlCl}_3 = 26.98 + 3(35.45) = 133.33\,\text{g/mol}$\text{Moles of AlCl}_3 = 0.125\,\text{L} \times 0.150\,\text{mol/L} = 0.01875\,\text{mol}$\text{Mass of AlCl}_3 = 0.01875\,\text{mol} \times 133.33\,\text{g/mol} = 2.4999\,\text{g} \approx 2.50\,\text{g}$

* *Answer:* **c. $2.50\,\text{g of AlCl}_3$**

• Question 6: How many grams of $\text{NaCl}$ are present in $50.0\,\text{mL}$ of a $0.552\,\text{M}$ solution? (Options: a. $0.640\,\text{g}$, b. $1.60\,\text{g}$, c. $2.45\,\text{g}$, d. $5.25\,\text{g}$)
  • Calculation:

$\text{Moles of NaCl} = 0.0500\,\text{L} \times 0.552\,\text{mol/L} = 0.0276\,\text{mol}$\text{Mass of NaCl} = 0.0276\,\text{mol} \times 58.44\,\text{g/mol} = 1.613\,\text{g} \approx 1.60\,\text{g}\n\n * *Answer:* **b. 1.60\,\text{g}**\n\n* **Equivalents Calculations:**\n * *Question 1:* Laboratory tests indicate a blood calcium level of 8.8\,\text{mEq/L}$. How many moles of calcium ion ($\text{Ca}^{2+}$) are present in$0.50\,\text{L} of blood?\n * *Calculation:*\n\n\text{Total mEq} = 0.50\,\text{L} \times 8.8\,\text{mEq/L} = 4.4\,\text{mEq} = 0.0044\,\text{Eq}$\text{Moles of Ca}^{2+} = \frac{0.0044\,\text{Eq}}{2\,\text{Eq/mol}} = 0.0022\,\text{mol} = 2.2 \times 10^{-3}\,\text{mol}$

* *Answer:* **$2.2 \times 10^{-3}\,\text{mol of Ca}^{2+}$ (or $0.0022\,\text{mol}$)**

• Question 2: What is the concentration of $\text{Mg}^{2+}$ in $\text{Eq/L}$ when $2.0\,\text{moles}$ of magnesium ions are dissolved in $10.\,\text{L}$ of solution? (Options: a. $2.0\,\text{Eq/L}$, b. $0.20\,\text{Eq/L}$, c. $0.40\,\text{Eq/L}$, d. $4.0\,\text{Eq/L}$)
  • Calculation:

$\text{Molarity} = \frac{2.0\,\text{mol}}{10.\,\text{L}} = 0.20\,\text{M}$\text{Equivalents/L} = 0.20\,\text{mol/L} \times 2\,\text{Eq/mol} = 0.40\,\text{Eq/L}\n\n * *Answer:* **c. 0.40\,\text{Eq/L}**\n * *Question 3:* Diagnostic charge match:\n * 1. In 1\,\text{mole}$of$\text{Fe}^{3+}$, there are: **c.$3\,\text{Eq}**\n * 2. In 2.5\,\text{moles}$of$\text{SO}_4^{2-}$, there are:$2.5 \times 2 =$**b.$5.0\,\text{Eq}**\n * 3. An IV bottle contains \text{NaCl}$. If$\text{Na}^+$is$34\,\text{mEq/L}$, the$\text{Cl}^-$is: **a.$34\,\text{mEq/L}** (due to electroneutrality)\n\n* **Dilution Calculations:**\n * *Question 1:* What is the final concentration when 0.50\,\text{L}$of$6.0\,\text{M } \text{HCl}$solution is diluted to a final volume of$1.0\,\text{L}?\n * *Calculation:*\n\nM_2 = \frac{M_1V_1}{V_2} = \frac{6.0\,\text{M} \times 0.50\,\text{L}}{1.0\,\text{L}} = 3.0\,\text{M}\n\n * *Answer:* **3.0\,\text{M}**\n * *Question 2:* What volume of a 2.00\%\,(\text{m/v})\,\text{HCl}$solution can be prepared by diluting$25.0\,\text{mL}$of$14.0\%\,(\text{m/v})\,\text{HCl} solution?\n * *Calculation:*\n\nV_2 = \frac{C_1V_1}{C_2} = \frac{14.0\% \times 25.0\,\text{mL}}{2.00\%} = 175\,\text{mL}\n\n * *Answer:* **175\,\text{mL}**\n * *Question 3:* What is the percent (% m/v) of a solution prepared by diluting 10.0\,\text{mL}$of$9.00\%\,\text{NaOH}$to$60.0\,\text{mL}?\n * *Calculation:*\n\nC_2 = \frac{C_1V_1}{V_2} = \frac{9.00\% \times 10.0\,\text{mL}}{60.0\,\text{mL}} = 1.50\%\n\n * *Answer:* **1.50\%\,(\text{m/v})**\n * *Question 4:* What is the molarity (M$) of a solution prepared by diluting$0.180\,\text{L}$of$0.600\,\text{M } \text{HNO}_3$to$0.540\,\text{L}?\n * *Calculation:*\n\nM_2 = \frac{M_1V_1}{V_2} = \frac{0.600\,\text{M} \times 0.180\,\text{L}}{0.540\,\text{L}} = 0.200\,\text{M}\n\n * *Answer:* **0.200\,\text{M}**\n * *Question 5:* A nurse needs to prepare 75\,\text{mL}$of a$0.35\%\,(\text{v/v})$solution from a stock solution of$1.2\%\,(\text{v/v})$medication. How much stock solution should the nurse use? (Options: a.$18.2\,\text{mL}$, b.$21.9\,\text{mL}$, c.$179\,\text{mL}$, d.$257\,\text{mL})\n * *Calculation:*\n\nV_1 = \frac{C_2V_2}{C_1} = \frac{0.35\% \times 75\,\text{mL}}{1.2\%} = 21.875\,\text{mL} \approx 21.9\,\text{mL}\n\n * *Answer:* **b. 21.9\,\text{mL}**\n * *Question 6:* What is the final volume of a solution prepared from diluting 25\,\text{mL}$of a$7.0\%\,(\text{m/v})$solution to a final concentration of$4.5\%\,(\text{m/v})$? Assume volumes add. (Options: a.$15\,\text{mL}$, b.$38\,\text{mL}$, c.$56\,\text{mL}$, d.$43\,\text{mL})\n * *Calculation:*\n\nV_2 = \frac{C_1V_1}{C_2} = \frac{7.0\% \times 25\,\text{mL}}{4.5\%} = 38.89\,\text{mL} \approx 39\,\text{mL} \text{ (closest given choice is 38 mL)}\n\n * *Answer:* **b. 38\,\text{mL}**\n * *Question 7:* How many milliliters of water must be added to 265\,\text{mL}$of a$1.75\,\text{M } \text{HCl}$solution to dilute it to$0.750\,\text{M } \text{HCl}$, assuming the volumes are additive? (Options: a.$353\,\text{mL}$, b.$114\,\text{mL}$, c.$202\,\text{mL}$, d.$618\,\text{mL})\n * *Calculation:*\n\nV_2 = \frac{M_1V_1}{M_2} = \frac{1.75\,\text{M} \times 265\,\text{mL}}{0.750\,\text{M}} = 618.33\,\text{mL}$\text{Volume of water added} = V_2 - V_1 = 618.33\,\text{mL} - 265\,\text{mL} = 353.33\,\text{mL} \approx 353\,\text{mL}$

* *Answer:* **a. $353\,\text{mL}$**

• Question 8: What is the concentration of a solution formed by diluting $3.7\,\text{mL}$ of a $6.2\,\text{M } \text{LiCl}$ solution to $135.0\,\text{mL}$? (Options: a. $0.170\,\text{M}$, b. $0.212\,\text{M}$, c. $0.226\,\text{M}$, d. $0.235\,\text{M}$)
  • Calculation:

$M_2 = \frac{M_1V_1}{V_2} = \frac{6.2\,\text{M} \times 3.7\,\text{mL}}{135.0\,\text{mL}} = 0.17007\,\text{M} \approx 0.170\,\text{M}$

* *Answer:* **a. $0.170\,\text{M}$**