Comprehensive Study Guide: Solution Concentrations and Units

Administrative Announcements and Logistics

• Discussion Section Relocations and Room Changes:

  • Yesterday Morning Section: Originally held in Kata Hay, this section has been officially moved to Chemistry Building 230. The teaching assistant (TA) unexpectedly failed to show up for yesterday morning's session.
  • Yesterday Afternoon Section: Currently assigned to Room 003 in the basement. This section of the building is experiencing air conditioner failure, turning the room into a "swamp." Two alternative rooms are available in the Chemistry Building, and an email will be distributed in the afternoon detailing the new room assignment to move students out of the basement.

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  • Homework Due Dates: Each section is accompanied by a problem set. Problem sets are strictly due one week (1 week) after the lecture finishes that specific section. For instance, if a section is completed on a Friday, its corresponding problem set will be due the following Friday. The first problem set is not due yet because the foundational material has not been fully covered in class.

• Office Hours:

  • Office hour schedules are published on D2L.
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• Classroom Etiquette:

  • Side conversations during lecture are discouraged; the instructor offered 50¢ to students engaged in chatter so they could buy coffee elsewhere.

Concentration Units Overview

• Molarity ($M$):

  • Represented by a capital letter $M$.
  • Defined as the number of moles of solute dissolved per liter of solution ($mol\,L^{-1}$).

• Mole Fraction:

  • A unitless quantity representing the ratio of the moles of a target component to the total moles of all components present in the mixture.
  • It is unitless because the unit of moles in the numerator cancels out with the unit of moles in the denominator ($\frac{\text{mole}}{\text{mole}}$).

• Molality ($m$):

  • Represented by a lowercase letter $m$.
  • Defined as the number of moles of solute per mass of solvent expressed in kilograms:     $\text{Molality } (m) = \frac{\text{moles of solute}}{\text{mass of solvent in } kg}$
  • Aqueous Solution Assumption Rule: For all calculations involving aqueous solutions in this course, $1\,L$ of water is defined as equivalent to $1\,kg$ of water (assuming a water density of $\rho = 1.00\,g\,mL^{-1}$ or $1.00\,kg\,L^{-1}$) unless explicitly stated otherwise in the prompt.
  • Example: A $1.00\,m$ aqueous sodium sulfate ($Na_2SO_4$) solution contains $142\,g$ of $Na_2SO_4$ (its molecular weight) dissolved in $1.00\,kg$ of water.

• Percent by Mass (Percent by Weight):

  • Defined as the ratio of the mass of the solute to the total mass of the solution, multiplied by $100$:     $\text{Percent by Mass} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100$
  • Commonly utilized for commercial packaged liquids (e.g., hydrogen peroxide, rubbing alcohol).
  • Conceptualized as "parts per hundred" (pph).
  • Sodium Sulfate Example: A solution prepared with $142\,g$ of sodium sulfate solute in $1000\,g$ of solvent has a total solution mass of $1142\,g$. The percent by mass is calculated as:     $\text{Percent Mass} = \frac{142\,g}{1142\,g} \times 100 = 12.4\%$

• Related Mass Ratio Concentration Units:

  • Parts per thousand (ppt): Calculated by multiplying the mass ratio by $1000$ ($10^3$).
  • Parts per million (ppm): Calculated by multiplying the mass ratio by $1\,000\,000$ ($10^6$).
  • Parts per billion (ppb): Calculated by multiplying the mass ratio by $1\,000\,000\,000$ ($10^9$).

Detailed Calculations for Concentration Units

• Exhaustive Worked Example: Glucose Solution:
  • Problem Prompt: A solution is prepared by dissolving $170.1\,g$ of glucose ($C_6H_{12}O_6$) in enough water to produce $1.00\,L$ ($1000\,mL$) of solution. The density of the resulting solution is $\rho = 1.062\,g\,mL^{-1}$. Express the concentration of the solution in molality ($m$), percent by mass, and parts per million (ppm).
  • Step 1: Calculate Moles of Solute ($C_6H_{12}O_6$)
  • The molecular mass of glucose ($C_6H_{12}O_6$) is calculated from the atomic weights ($C = 12.011\,g\,mol^{-1}$, $H = 1.008\,g\,mol^{-1}$, $O = 15.999\,g\,mol^{-1}$) as approximately $180.16\,g\,mol^{-1}$ (minor variations in decimal rounding from different periodic tables do not affect calculation accuracy).     $\text{Moles of glucose} = \frac{170.1\,g}{180.16\,g\,mol^{-1}} = 0.9442\,mol$
  • Step 2: Determine Total Solution Mass
  • Using the given density of $1.062\,g\,mL^{-1}$ for a $1000\,mL$ solution:     $\text{Mass of solution} = 1.062\,g\,mL^{-1} \times 1000\,mL = 1062\,g$
  • Step 3: Determine Solvent Mass (Water)
  • Subtract the mass of the solute from the total solution mass:     $\text{Mass of water} = 1062\,g - 170.1\,g = 891.9\,g = 0.892\,kg$
  • Step 4: Calculate Molality ($m$)$\text{Molality } (m) = \frac{0.9442\,mol}{0.892\,kg} = 1.06\,m$
  • Step 5: Calculate Percent by Mass$\text{Percent Mass} = \frac{170.1\,g}{1062\,g} \times 100 = 16.0\%$
  • Step 6: Calculate Parts per Million (ppm)$\text{ppm} = \frac{170.1\,g}{1062\,g} \times 10^6 = 160\,169\,ppm \approx 1.60 \times 10^5\,ppm$

Selecting the Appropriate Concentration Unit

• Molarity ($M$):

  • Primary Applications: Preferred for volumetric procedures such as titrations and gravimetric analysis.
  • Advantages: Measuring liquid volumes using calibrated glassware is simpler and more convenient than weighing solvent masses.
  • Limitations: Temperature-dependent because liquid volume expands or contracts with temperature changes. Solvents with low boiling points may evaporate during laboratory operations, inadvertently altering the volume and concentration.

• Mole Fraction:

  • Primary Applications: Preferred when working with gas mixtures and measuring colligative properties like vapor pressure (e.g., Raoult's Law).

• Molality ($m$):

  • Primary Applications: Required for experiments conducted across wide temperature ranges.
  • Advantages: Completely temperature-independent because solvent mass remains constant regardless of temperature fluctuations.

• Percent by Mass:

  • Primary Applications: Standard in commercial chemical manufacturing and consumer liquid labeling.
  • Advantages: Temperature-independent and does not require knowledge of the molar mass of the solute.

Converting Between Concentration Units

• Conversion Type 1: Molality to Molarity

  • Problem: Express a $0.396\,m$ aqueous glucose ($C_6H_{12}O_6$) solution at $25\,^\circ\text{C}$ in Molarity ($M$), given a solution density of $\rho = 1.16\,g\,mL^{-1}$.
  • Calculation Step-by-Step:
  1. Assume Solvent Basis: Assume exactly $1.00\,kg$ ($1000\,g$) of solvent water, containing $0.396\,mol$ of glucose.
  2. Mass of Solute:$\text{Mass of glucose} = 0.396\,mol \times 180.16\,g\,mol^{-1} = 71.34\,g$
  3. Total Mass of Solution:$\text{Mass of solution} = 1000\,g + 71.34\,g = 1071.34\,g$
  4. Volume of Solution:$\text{Volume of solution} = \frac{1071.34\,g}{1.16\,g\,mL^{-1}} = 923.57\,mL = 0.9236\,L$
  5. Calculate Molarity:$\text{Molarity } (M) = \frac{0.396\,mol}{0.9236\,L} = 0.429\,M$

• Conversion Type 2: Mass Percent to Molarity and Molality

  • Problem: Commercial rubbing alcohol is a mixture of isopropyl alcohol ($C_3H_7OH$) and water that is $70.0\%$ isopropyl alcohol by mass, with a solution density of $\rho = 0.790\,g\,mL^{-1}$ at $20\,^\circ\text{C}$. Calculate both the molarity ($M$) and molality ($m$).
  • Calculation Step-by-Step:
  1. Molar Mass: Isopropyl alcohol ($C_3H_7OH$) molar mass = $60.09\,g\,mol^{-1}$.
  2. Mass of 1 L Solution:$\text{Mass of } 1\,L \text{ solution} = 0.790\,g\,mL^{-1} \times 1000\,mL = 790\,g$
  3. Mass of Solute:$\text{Mass of isopropyl alcohol} = 0.700 \times 790\,g = 553\,g$
  4. Moles of Solute and Molarity:$\text{Moles of alcohol} = \frac{553\,g}{60.09\,g\,mol^{-1}} = 9.20\,mol$$\text{Molarity } (M) = \frac{9.20\,mol}{1.00\,L} = 9.20\,M$
  5. Mass of Solvent and Molality:$\text{Mass of water} = 790\,g - 553\,g = 237\,g = 0.237\,kg$$\text{Molality } (m) = \frac{9.20\,mol}{0.237\,kg} = 38.8\,m \approx 39\,m$

In-Class Practice Problems and Calculations

• Practice Problem 1: Mass Percent Calculation

  • Prompt: Determine the percent by mass of $KCl$ in a solution prepared by dissolving $3.01\,g$ of $KCl$ in $72.5\,g$ of water.
  • Detailed Solution:$\text{Total Solution Mass} = 3.01\,g + 72.5\,g = 75.51\,g$$\text{Percent Mass } KCl = \frac{3.01\,g}{75.51\,g} \times 100 = 3.986\% \approx 3.99\%$
  • Correct Option: Option C ($3.99\%$).

• Practice Problem 2: Molality of Biphenyl Solution

  • Prompt: Calculate the molality ($m$) of a solution containing $6.44\,g$ of biphenyl ($C_{12}H_{10}$) dissolved in $80.0\,g$ of benzene.
  • Detailed Solution:
  • Molar mass of biphenyl ($C_{12}H_{10}$): $12 \times 12.011 + 10 \times 1.008 = 154.21\,g\,mol^{-1}$.
  • Moles of biphenyl:       $\text{Moles} = \frac{6.44\,g}{154.21\,g\,mol^{-1}} = 0.04176\,mol$
  • Mass of solvent in kilograms:       $\text{Benzene mass} = 80.0\,g = 0.0800\,kg$
  • Molality calculation:       $\text{Molality } (m) = \frac{0.04176\,mol}{0.0800\,kg} = 0.522\,m$

• Practice Problem 3: Sucrose Molality to Molarity Conversion

  • Prompt: At $20\,^\circ\text{C}$, a $0.258\,m$ aqueous solution of sucrose ($C_{12}H_{22}O_{11}$) has a density of $\rho = 1.0173\,g\,mL^{-1}$. Calculate the molarity ($M$) of this solution.
  • Detailed Solution:
  • Assume basis of $1.00\,kg$ ($1000\,g$) solvent water containing $0.258\,mol$ sucrose.
  • Molar mass of sucrose ($C_{12}H_{22}O_{11}$): $342.30\,g\,mol^{-1}$.
  • Mass of sucrose solute:       $\text{Mass of sucrose} = 0.258\,mol \times 342.30\,g\,mol^{-1} = 88.31\,g$
  • Total mass of solution:       $\text{Total mass} = 1000\,g + 88.31\,g = 1088.31\,g$
  • Volume of solution using density:       $\text{Volume} = \frac{1088.31\,g}{1.0173\,g\,mL^{-1}} = 1069.80\,mL = 1.0698\,L$
  • Molarity calculation:       $\text{Molarity } (M) = \frac{0.258\,mol}{1.0698\,L} = 0.241\,M$
  • Correct Option: Option B ($0.241\,M$).

• Practice Problem 4: Sulfuric Acid Mass Percent Conversion

  • Prompt: At $25\,^\circ\text{C}$, an aqueous solution of $15.0\%$ sulfuric acid ($H_2SO_4$) by mass has a density of $\rho = 1.178\,g\,mL^{-1}$. Calculate the molarity ($M$) and molality ($m$) of this solution.
  • Detailed Solution:
  • Molar mass of $H_2SO_4$: $2 \times 1.008 + 32.06 + 4 \times 15.999 = 98.08\,g\,mol^{-1}$.
  • Assume basis of $1.00\,L$ ($1000\,mL$) of solution.
  • Mass of $1.00\,L$ solution:       $\text{Mass of solution} = 1.178\,g\,mL^{-1} \times 1000\,mL = 1178\,g$
  • Mass of solute ($H_2SO_4$):       $\text{Mass of } H_2SO_4 = 0.150 \times 1178\,g = 176.7\,g$
  • Moles of solute ($H_2SO_4$):       $\text{Moles of } H_2SO_4 = \frac{176.7\,g}{98.08\,g\,mol^{-1}} = 1.801\,mol$
  • Molarity calculation:       $\text{Molarity } (M) = \frac{1.801\,mol}{1.00\,L} = 1.80\,M$
  • Mass of solvent water:       $\text{Mass of water} = 1178\,g - 176.7\,g = 1001.3\,g = 1.0013\,kg$
  • Molality calculation:       $\text{Molality } (m) = \frac{1.801\,mol}{1.0013\,kg} = 1.80\,m$

Questions and Discussion

• Audience Question on Solvent Assumptions:

  • Question: In an aqueous solution, are we always going to assume that there is a kilogram of water unless specified otherwise?
  • Answer: Yes. Unless stated otherwise in the specific question prompt, assume $1.00\,kg$ (or $1.00\,L$) of water solvent to streamline calculations.

• Audience Question on Unit Dimensions:

  • Question: Is mole fraction unitless?
  • Answer: Yes, mole fraction is unitless because the mole unit in the numerator cancels out with the mole unit in the denominator.