Lesson 2 - Solutions and Types of Concentration PDF

# GENERAL CHEMISTRY II ## **Chapter 2** Solutions and Their Properties ### WEEK 3-4 # Most Essential Learning Competencies 1. Use different ways of expressing concentration of solutions: percent by mass, mole fraction, molarity, molality, percent by volume, percent by mass, ppm. 2. Perform sto...

# GENERAL CHEMISTRY II ## **Chapter 2** Solutions and Their Properties ### WEEK 3-4 # Most Essential Learning Competencies 1. Use different ways of expressing concentration of solutions: percent by mass, mole fraction, molarity, molality, percent by volume, percent by mass, ppm. 2. Perform stoichiometric calculations for reactions in solution. 3. Describe the effect of concentration on the colligative properties of solutions. 4. Differentiate colligative properties of nonelectrolyte solutions and of electrolyte solution. 5. Calculate the boiling point elevation and freezing point depression from the concentration of a solute in a solution. 6. Calculate molar mass from colligative property data. 7. Describe laboratory procedures in determining concentration of solution. # General Chemistry 2 – Senior High School (STEM) ## **Section 2.1: Types of Solutions** **EQ:** What types of solutions are encountered everyday? # Introduction Identify the solute and solvent in each of the following solutions, explain your answers. 1. 25 grams of salt dissolved in 95 mL of water. 2. 25 mL of water mixed with 75 mL of isopropyl alcohol. 3. Tincture of iodine prepared with 0.20 gram of iodine and 20.0 ml of ethanol. # Solutions - Solutions are mixtures of two or more substances distributed throughout a single phase. - Atoms, ions, or molecules are thoroughly mixed in a solution such that each part of the mixture has uniform composition and properties. - A solution consists of a solute and a solvent. - The **Solute** is the substance dissolved in a solution and is usually present in smaller amount; **solvent** is the dissolving medium. # Types of Solutions A table describing types of solutions as categorized by state of matter. | SOLUTE | SOLVENT | SOLUTION | EXAMPLE | |---|---|---|---| | GAS | GAS | GAS | Oxygen in Nitrogen | | GAS | LIQUID | LIQUID | Carbon Dioxide in water | | GAS | SOLID | SOLID | Hydrogen in Palladium | | LIQUID | LIQUID | LIQUID | Ethanol in water | | LIQUID | SOLID | SOLID | Mercury in silver | | SOLID | LIQUID | LIQUID | Salt in water | | SOLID | SOLID | SOLID | Copper in tin (bronze) | # Type of Solution - A solution can be classified as unsaturated, saturated, or supersaturated depending on the quantity of solute present in it. ## Unsaturated Solution - more solute dissolves ## Saturated Solution - no more solute dissolves ## Supersaturated Solution - becomes unstable, crystals form # Types of Solution - **Unsaturated Solution** contains less solute than solvent's capacity to dissolve. This means that the solvent can still dissolve more solute. - **Saturated Solution** contains the maximum amount of solute that the solvent can dissolve at a certain temperature. - **Supersaturated solution** contains more dissolved solute than is present in a saturated solution. This type of solution is unstable such that the excess solute can crystallize in the solution by adding a "seed crystal" (a process called seeding) or by scratching the sides of the container. # General Chemistry 2 ## **Section 2.2: Energy of Solution Formation** **EQ:** How do liquid solution forms? # Energy of Solution - For two substances to form a solution, they must have the same nature in terms of Polarity. - The formation of liquid solution takes place in three steps: 1. Overcoming the intermolecular forces in the solvent to give room for the solute. 2. Separating the solute into its individual components. 3. Allowing the solute and solvent to interact to form the solution. # Enthalpy of Solution - It is the enthalpy change associated with the formation of the solution. - It is equal to the sum of the $ΔH$ values for the three steps. - A positive enthalpy of solution signifies an endothermic process, while negative indicates an exothermic process. - The process is exothermic, if more energy is released when new bonds form than is used when bonds are broken. - The process is endothermic, if more energy is required (used) when bonds are broken than is released when new bonds are formed. # General Chemistry 2 ## **Section 2.3: Concentration of Solutions** **EQ:** Why should solutions be expressed in a correct concentration units? # Concentration of Solution - It is a measure of the amount of solute in a given amount of solvent or solution. - One type of solution may be prepared and expressed in different concentrations. - Example: - A cup of hot water with one teaspoon of coffee is altogether different solution than a cup of water with a few teaspoon of coffee. - Medications are also solutions of substances, A 5 mL dose of a certain concentration of medicine might cure a sick person, but a 5 mL dose of a higher concentration might already harmful to the person. # Concentration of Solution 1. Percent by Mass 2. Percent by Volume 3. Percent by Mass-Volume 4. Mole Fraction 5. Molality 6. Molarity 7. Parts per Million # Percent by Mass and Volume Percent by Mass = (Mass of Solute / Mass of Solution) * 100 Percent by Volume = (Volume of Solute / Volume of Solution) * 100 # Sample Problem #1 - A saline solution with a mass of 355 g has 36.5 g of NaCl dissolved in it. What is the mass percent concentration of the solution? - ** Solution** - Given: - Mass of Solute = 36.5 grams - Mass of Solution = 355 grams - Formula: Percent by mass = (mass of solute / mass of solution) * 100 - Percent by mass = (36.5 grams / 355 grams) * 100 = 10.28% # Sample Problem #2 - A wine contains 12% alcohol by volume. Calculate the volume (in mL) of alcohol in 350 mL of the wine. - **Solution:** - Given: - Percent by Volume = 12% - Volume of Solution = 350 mL - Volume of solute = x - Formula: % by volume = (volume of solute / volume of solution) * 100 - 12% = (x / 350 mL) * 100 - (12 / 100) * (350 mL) = x - 42 mL = x # Percent by Mass - Volume Percent by Mass – Volume = (mass of solute in grams / volume of solution in mL) * 100 # Sample Problem #1 - What is the concentration in percent by mass/volume of 150 mL of solution containing 30 g of solute? - **Solution** - Given: - volume of solution (mL) : 150 ml - mass of solute (g) : 30 grams - Find: % by mass-volume - Formula: % by mass-volume = (mass of solute in g / volume of solution in ML) * 100 - % by mass-volume = (30 grams / 150 mL) * 100 = (0.20) * 100 - = 20% # Sample Problem #2 - A 50 mL of 12% by mass-volume solution was used in an experiment. How many grams of solute does the solution contain? - **Solution** - Given: - volume of solution in mL = 50 mL - % by mass-volume = 12% - Find: Mass of solute in grams = ? - Formula: % by mass-volume= (mass of solute in g / volume of solution in ML) * 100 - 12% = (x / 50 mL) * 100 - (12 / 100) * (50ml) = x - 6 g = x # Mole Fraction - The mole fraction (x) of a component in a solution is equal to the number of moles of the component divided by the total number of moles of all the components present. - The sum of all of the mole fractions of all components in all solution will always be equal to one. # Mole Fraction (X) $X_{solute}$ = (Moles of solute / Total moles of solution) $X_{solvent}$ = (Moles of solvent / Total moles of solution) Where: $X_{solute} + X_{solvent}$ = 1 # Sample Problems 1. 25g of NaF is mixed with 200 g of H₂O. What is the mole fraction of NaF in the solution? - **Step 1:** - Given: - solute: 25 g of NaF - solvent: 200 g of H₂O - Find: mole fraction of NaF # Mole Fraction (X) $X_{solute}$ = (Moles of solute / Total moles of solution) $X_{solvent}$ = (Moles of solvent / Total moles of solution) Where: $X_{solute} + X_{solvent}$ = 1 # Solution - Formula: $X_{solute}$ = (Moles of solute / Total moles of solution) - Step 2: Convert the given mass of solute and solvent into moles using molar mass. - Given: - Mass to mole - solute: 25 g of NaF - Molar Mass - Na : 1 x 23 = 23 g/mole - F : 1 × 19 = 19 g/mole - Total Molar Mass: 42 g/mole - Mass to mole: 25 g of NaF * (1 mole / 42 g) = 0.60 moles of NaF - Moles of solute: 0.60 mole of NaF - Mass to mole - solvent: 200 g of H₂O - Molar Mass - H: 2 x 1 = 2 g/mole - O : 1 x 16 = 16 g/mole - Total Molar Mass: 18 g/mole - Mass to mole: 200 g of H₂O * (1 mole / 18 g) = 11.11 moles of H₂O - moles of solvent: 11.11 moles of H₂O - moles of solute + moles of solvent = Total Moles of Solution - Moles of solute = 0.60 moles of NaF - Moles of solvent = 11.11 moles of H₂O - Total Moles of Solution = 11.71 moles - Moles of solute = 0.60 moles of NaF - Total Moles of Solution = 11.71 moles - $X_{solute}$ = (Moles of solute / Total moles of solution) = 0.60 moles of NaF / 11.71 moles = 0.05 (Final Answer) - Moles of solvent = 11.11 moles of H₂O - Total Moles of Solution = 11.71 moles - $X_{solute}$ = (Moles of solute / Total moles of solution) = 11.11 moles / 11.71 moles = 0.95 # Mole Fraction (X) $X_{solute}$ = (Moles of solute / Total moles of solution) $X_{solvent}$ = (Moles of solvent / Total moles of solution) Where: $X_{solute} + X_{solvent}$ = 1 $X_{solute}$ = 0.05 $X_{solvent}$ = 0.95 Total = 1 # Sample Problems - Calculate the mole fraction of each component of a solution containing 65 g of ethanol (C₂H₅OH) in 350 g of water? # Molality - The molality of a solution is the number of moles of solute per kilogram of solvent. - Molality= (number of moles of solute / weight of solvent in kg) # Sample Problems - Calculate the molality of a solution containing 16.5 g of dissolved naphthalene (C₁₀H₈) in 0.0543 Kg benzene (C₆H₆). # Solution - Calculate the molality of a solution containing 16.5 g of dissolved naphthalene (C₁₀H₈) in 0.0543 Kg benzene (C₆H₆). - Given: - Solute: 16.5 g C₁₀H₈ - Solvent: 54.3 g C₆H₆ - Find: molality - Molality= (number of moles of solute / weight of solvent in kg) - STEP 2: Convert the mass of solute into moles using its molar mass. - Solute: 16.5 g C₁₀H₈ - C : 10 × 12 = 120 g/mole - H: 8 x 1= 8g/mole - Total Molar Mass = 128 g/mole - 16.5 g C₁₀H₈ x (1 mole / 128 g) = 0.132 moles of C₁₀H₈ - STEP 3: Calculate the Molality using the formula. - Molality = (number of moles of solute / weight of solvent in kg) - m = (0.132 moles / 0.0543 Kg) - m = 2.49 moles/Kg # Sample Problem - 10g of NaOH is dissolved in 500 g water. What is the molality of solution? # Molarity - Molarity of solution is the number of moles of solute per liter of solution. It is especially useful in doing stoichiometric calculations involving solution. - M = (Moles of Solute / Liters of Solution) # Sample Problems - Determine the molarity of a solution containing 2.40 g of sodium chloride (NaCl) in 40.0 mL of solution. # Solution - Determine the molarity of a solution containing 2.40 g of sodium chloride (NaCl) in 40.0 mL of solution. - STEP 1 - Given - Solute: 2.40 g of NaCl - Solution: 40 mL ------ 0.04 L - Find: Molarity - STEP 2: Convert the given mass of solute into moles using molar mass. - Solute: 2.40 g of NaCl - Na : 1 x 23 = 23 g/mole - Cl : 1 × 35 = 35 g/mole - Total Molar Mass of solute: 58 g/mole - 2.40 g of NaCl * (1 mole / 58 grams) = 0.04 moles - STEP 3: Calculate the molarity using the formula. - M = (Moles of Solute / Liters of Solution) - M = (0.04 moles / 0.04 L) - M = 1 mole/L # Sample Problems - Determine the molar concentrations of a solution that contains 25g of potassium hydroxide (KOH) in 250 mL of solution. # Sample Problem - What is the molarity of 0.25 mol of NaCl in 300 mL of solution? # Parts Per Million/Billion - Parts per million (ppm) expresses the number of parts of solute per one million/billion parts of solution. - ppm = (Mass of Solute / Mass of Solution) * 10⁶ - ppb = (Mass of Solute / Mass of Solution) * 10⁹ # Sample Problem - 25 grams of sodium chloride is dissolved in 100 grams of water, What is the concentration of the sodium in parts per million (ppm)? - STEP 1 - Given - Solute : 25 grams NaCl - Solvent : 100 grams water - Solution : 125 grams - Find: ppm - STEP 2: Calculate the ppm using the formula. - ppm = (Mass of Solute / Mass of Solution) * 10⁶ - ppm = (25 grams / 125 grams) * 10⁶ - ppm = 200,000