General Chemistry 2 Module 5 PDF

SHS General Chemistry 2 Quarter 2: Week 5 - Module 5 Nonelectrolyte and Electrolyte Solutions STEM – General Chemistry 2 Grade 11/12 Quarter 3: Week 5 - Module 5: Nonelectrolyte and Electrolyte Solutions First Edition, 2021 Copyright © 2021...

SHS General Chemistry 2 Quarter 2: Week 5 - Module 5 Nonelectrolyte and Electrolyte Solutions STEM – General Chemistry 2 Grade 11/12 Quarter 3: Week 5 - Module 5: Nonelectrolyte and Electrolyte Solutions First Edition, 2021 Copyright © 2021 La Union Schools Division Region I All rights reserved. No part of this module may be reproduced in any form without written permission from the copyright owners. Development Team of the Module Author: Femy B. Nuesca, MT- II Editor: SDO La Union, Learning Resource Quality Assurance Team Illustrator: Ernesto F. Ramos Jr., P II Management Team: Atty. Donato D. Balderas, Jr. Schools Division Superintendent Vivian Luz S. Pagatpatan, Ph.D Assistant Schools Division Superintendent German E. Flora, Ph.D, CID Chief Virgilio C. Boado, Ph.D, EPS in Charge of LRMS Rominel S. Sobremonte, Ed.D, EPS in Charge of Science Michael Jason D. Morales, PDO II Claire P. Toluyen, Librarian II  General Chemistry 2 Quarter 3: Week 5 - Module 5 NONELECTROLYTE AND ELECTROLYTE SOLUTIONS  TARGET The properties of a solution depend on the relative amounts of the components of a solution, the solute and the solvent. Solution properties like the colligative properties are properties which depend on the amount or concentration of the solute and are determined by the number of solute particles dissolved in a fixed quantity of solvent. Note that colligative properties depend on the total concentration of solute and is not depended unto the nature or identity of the solute particles. This module will give concrete evidence why adding salt into the boiling water will increase its temperature and give justice why putting salts affect how water freezes and effectively lowers the freezing/melting point of water, and why the ice vendor adds salts to the ice where the containers of the ice cream are placed and suddenly lowers its melting point. (Note: Melting point=Freezing point) This module involves problem solving related with boiling point elevation and freezing point depression. It also involves calculations of molar mass from the colligative properties of boiling point elevation and freezing point Thereby, after going through this module, you are expected to: 1. Differentiate the colligative properties of nonelectrolyte solutions and electrolyte solutions 2. Calculate the boiling point elevation and freezing point depression from the concentration of a solute in a solution 3. Calculate the molar mass from the colligative property data. Jumpstart Knowledge Checkpoint!!! Arrange the following according to increasing boiling point and freezing point. 1m AlCl3, 1m CaCl2, 1m NaCl, 1m C6H12O6 Increasing Boiling Point ________________________________________________________________ Increasing Freezing Point  DISCOVER Electrolyte vs. Non-electrolyte From the previous lesson, you were task to describe the effect of the concentration on the colligative properties of solution. In this module, you are expected to differentiate the colligative properties between the effects that an electrolyte and that of a non-electrolyte solution. In a better journey in solving problems related with colligative properties, let as again define its meaning. Colligative properties are properties of a solution that depend only on the colligative effect of the concentration of solute particles present, that is into the number of the solute.Colligative properties has a direct relationship to the number of solute particles. The colligative properties are very useful for characterizing the nature of a solute after it is dissolved in a solvent and for determining the molar mass of a substance. The concentration of a solute that does not have a vapor pressure of its own. The solution has an effect on the colligative properties of solution, the effect would depend on the ratio of the number of particles of solute in the solution and not on the identity of the solute. However, it is necessary to consider whether the solute is an electrolyte or non-electrolyte. Ionic compounds like sodium chloride (NaCl), are strong electrolytes that dissociate into ions. Dissociation means the addition of a solvent or of energy in the form of heat causes molecules or crystals of the substance to break up into ions (electrically charged particles. Thus, when a ionic compound is dissolved in a solution results in a larger number of dissolved particles. But other compounds like molecular compounds sucrose does not dissociate into ions. Let us take a look to the equation below. NaCl(s) → Na+(aq) + Cl-(aq) 1 = Na ion and 1 = Cl ion (2 dissolved particles) C12H22O12(s) → C12H22O12(aq) 1 dissolved particle Ca + Cl2 → CaCl2 1 Ca and 2 Cl ions (3 dissolved particles) ( look at the arrow for the number of ions or particles )  Sodium chloride (NaCl) dissociates into two particles and CaCl 2 dissociates into 3 particles, while sucrose (C12H22O12) does not dissociate. Thus, equal concentrations of each solution will result in twice as many dissolved particles as in the case of Sodium chloride (NaCl) as compared to the sucrose (C12H22O12). Refer to the following table showing the difference between an electrolyte and non- electrolyte. Table 1: Electrolytes vs. Non-electrolytes Colligative Properties Electrolytes Non-electrolytes Colligative properties of electrolytes are Colligative properties of non-electrolytes the physical properties of electrolytic are the physical properties of non- solutions that depend on the amount of electrolytic solutions that depend on the solutes regardless the nature of solutes amount of solutes regardless the nature of solutes Solutes Electrolytes provide more solutes to the Non-electrolytes provide low solutes to solution via dissociation; hence, the the solution since there is no colligative properties are considerably dissociation; hence, the colligative changed. properties are not considerably changed. Effect on Colligative Properties The effect of electrolytes on colligative The effect of non-electrolytes on properties is very high compared to non- colligative properties is very low electrolytes. compared to electrolytes. Boiling Point Elevation What are non-volatile solutes again? Yes, you are correct! These are solutes that does not have a vapor pressure on its own. The addition of non-volatile solute like NaCl lowers the vapor pressure of the solution. Thus, increases the boiling point. The boiling point is defined as the temperature at which the saturated vapor pressure of a liquid is equal to the surrounding atmospheric pressure. As far as boiling water is concerned, the vapor pressure is directly related to boiling point. By definition, water boils at the temperature where the atmospheric pressure equals the vapor pressure. Furthermore, when the only volatile component of a solution is the solvent, then the vapor pressure of the solution is less than the vapor pressure that the solvent has at any given temperature. At the temperature at which the pure solvent normally boils, the vapor pressure of the solution is still not equal to the atmospheric pressure. So to make the vapor pressure of the solution come up to atmospheric pressure, we have to increase the temperature of the solution further. The presence of a non-volatile solute thus elevates the boiling point of the solution. The most common application of this property of solutions occurs in the use of permanent-type anti-freezers. These protect the liquid in a vehicle’s cooling system not just from freezing but also from boiling over. These products are based on either propylene glycol or ethylene glycol, which are high-boiling, nearly non- volatile, water soluble liquids, as they elevate the boiling point as well as lower the freezing point when dissolved in water. For dilute solution, the elevation of the boiling point is directly proportional to the molal concentration of the solute. ΔB = KB m ΔB = KB moles of solute kg of solution where, ΔB= boiling point elevation (K) KB = molal point boiling point elevation constant m = molality of solute The molal boiling point elevation constant (K mol -1kg). KB has a specific value depending on the identity of the solvent. Sample Problem 1: Calculate the boiling point of a solution containing 5g sucrose C12H22O11 (molar mass = 342.3 g) dissolved in 180 g of water H 2O. The mass of solute (5g ) is given which is dissolved in 180 g of solvent, water H2O. (0.512 – Kb of H2O) Given: 5g - grams of sucrose (C12H22O11) 342.3 g – molar mass of sucrose (C12H22O11) 180g – grams of water (H2O) 0.512 – Kb of H2O Step 1: Solve for the molality of the solution, the unit of which I mole solute/kilogram solvent. Using the dimensional analysis, the first factor converts mass to mole, then divide with the mass of solvent, and the last factor converts the mass in grams to kilograms. Molality (m) = mole solute kg solvent Molality (m) = 5 g C12H22O11 x 1 mole______ x 1____ x 1000 g 342.3 g C12H22O11 180 g H2O 1 kg = 5 x 1 mole x 1 x 1000 342.3 x 180 x 1 kg = 5000 mole 61614 kg = 0.081 mole/kg Step 2: Solve for boiling point elevation ΔTb = Kb x m = 0.512OC x 0.081 m 1m = 0.041OC Step 3: Solve for boiling point of the solution Boiling point of solution = boiling point of solvent + ΔTb 100OC + 0.041OC = 100.041OC ANSWER: 100.041OC This elevates its boiling point temperature by 0.041 OC from 100OC to 100.041OC. Sample Problem 2: What is the new boiling point of a solution prepared by adding 85.0 g of Sodium acetate to 356 mL of water? The boiling point constant for water is 0.52 °C/m. Step 1: Determine the molality of the NaCl solution. Molality of NaCl = 85.0 g / 58.443 g/mol 0.356 kg = 4.0854 m Step 2: Solve for boiling point elevation Δt = Kb m x = (0.52 °C kg mol¯1) (4.0854mol/kg) x = 2.124°C Step 3: Solve for boiling point of the solution Boiling point of solution = boiling point of solvent + ΔTb 100OC + 2.2124OC = 102.2124OC ANSWER: 102.2124OC This elevates its boiling point temperature by 2.2124OC from 100OC to 102.2124OC. Freezing Point Elevation The addition of non-volatile solute into pure solvent lowers the vapor pressure of the solvent resulting into lowering the freezing point of the solution s compared with the pure solvent. The freezing point depression is the difference in temperature between the freezing point of a pure solvent and that of a solution, a pure solvent and with the added non-volatile solute. At a given temperature, if a substance is added to a solvent like water, the solute-solvent interactions prevent the solvent from going into solid phase, requiring the temperature to decrease further before the solution will solidify. The magnitude of the freezing point depression is directly proportional to the molality of the solution. ΔTf = Kf m where; ΔTf = freezing point depression m = molality of solute Kf = molal freezing point depression constant That is equal to the change in freezing point for a one (1) molal solution of a non-volatile molecular solute. Sample Problem 1: Determine the freezing point of a solution of 10.0g Urea, CO(NH2)2, in 2.50 x 102 g of water. Given: 10.0 CO(NH2)2 2.50 x 102 g of water. STEPS: Step 1: Find for mol CO(NH2)2 g CO(NH2)2 → mol CO(NH2)2 1 mol CO(NH2)2 = 60.06 g CO(NH2)2 Construct the setup and calculate the final answer. 10 g CO(NH2)2 x 1 mol CO(NH2)2 = 0.167 mol CO(NH2)2 60.06 g CO(NH2)2 Step 2: Now calculate the molality of the solution by using its defining equation. m≡ mol solute = 0.167 mol CO(NH2)2 x 1000 g H2O = 0.067 m CO(NH2)2 kg solvent 25.0 x 102 g H2O 1 kg H2O Step 3: Find the freezing point depression by substitution into the equation ΔTf = Kf m. Kf for water is 1.86O C/m. ΔTf = Kf m = 1.86O C x 0.067 m = 0.125O C m Step 4: The freezing point depression is 1.49O C. The normal freezing point of water is 0O C. State the freezing point of the solution. 0O C – 0.125O C = -0.125O C ANSWER: -0.125O C This decreases by 0.125OC from 0OC to -0.125OC. Sample Problem 2: What is the freezing point depression when 62.5 g of toluene (C 7H8) is dissolved in 477 g of Naphthalene? The freezing point constant for naphthalene is 7.00 °C/m. Step 1: Find for mol C7H8 g C7H8→ mol C7H8 1 mol C7H8= 92.1402 g/mol C7H8 Construct the setup and calculate the final answer. 62.5 g C7H8 x 1 mol CO(NH2)2 = 0.678 mol C7H8 92.1402 g C7H8 Step 2: Now calculate the molality of the solution by using its defining equation. m≡ mol solute = 0.678 mol C7H8 = 1.421 m kg solvent 0.477 kg C10H8 Step 3: Find the freezing point depression by substitution into the equation ΔTf = Kf m. Kf for water is 1.86O C/m. ΔTf = Kf m = 7.00O C x 1.421 m = 9.95O C M ANSWER: 9.95O C 9.95O C is the amount the freezing point is depressed. How to Calculate the Molar Mass of a Solute from Freezing Point Depression or Boiling Point Elevation Data Step 1: Calculate molality from m= ΔTf/Kf or m = ΔTb/Kb. Express as mol solute/kg solvent. Step 2: Using molality as a conversion factor between moles of solute and kilograms of solvent, find the number of moles of solute. Step 3: Use the defining equation for molar mass, MM ≡ g/mol, to calculate the molar mass of the solute. Sample Problem 1: What is the molecular mass of an organic compound if 15.0 g of the compound is dissolved in 220.0 g of carbon tetrachloride raising the boiling point of 84.36OC? Boiling point of carbon tetrachloride is 76.72OC, Kb for CCl4 = 5.02OC/m. Given: 15.0 g - g of the compound 220.0 g CCl4 – mass of CCl4 84.36OC – raised boiling point of CCl4 76.72OC – normal boiling point of CCl4 5.02OC/m - Kb for CCl4 Step 1: Solve for ΔTb BP solution = BP solvent + ΔTb ΔTb = 84.36 – 76.72 = 7.64OC Step 2: Solve for the molality of the solution ΔTb = Kb x m 7.64OC = 5.02OC x m 1m Molality = 7.64OC x 1m___ = 1.522 m or 1.522 mole solute/kg water 5.02OC Step 3: Solve for the number of moles of the unknown from the molality Molality ( m ) = no. of moles Kg solvent No. of moles = 1.522 mole/kg x 220g x 1kg__ = 0.3348 mole 1 1000g Step 4: Solve for the molar mass ( MM ) Molar mass = mass solute = 15.00g___ = 44.80g/mol No. of moles 0.3348 moles ANSWER: 44.80g/mol NOTE : Don’t get confused with molality ( m ) and Molar Mass ( MM) Sample Problem 2: How many grams of Pyrazine (C4H4N2) would have to be dissolved in 2 kg of carbon tetrachloride to lower the freezing point by 5.6 °C? The freezing point constant for carbon tetrachloride is 30.0°C/m. Solution: Δt = Kf m 5.6 °C = (30.0 °C kg mol¯1) x___ 2kg 5.6 °C = (15. °C mol¯1) (x) x = 0.37 mol mass in g= 0.37 x 80.0896 g/mol = 29.633 g ANSWERS: 29.633 g EXPLORE The following activities will determine whether you understand the concepts discussed. Have fun! Activity 1: Solve the following problems: Show your any pertinent solutions. 1) What is the normal boiling point of a 3.45 mol solution of KBr that has a density of 1.80 g/ml? (KB for H2O is 0.512OC kg/mole) 2) Eugenol is the active ingredient in the oil of cloves used to relieve toothache. Calculate the boiling point of a solution in which 0.20 grams of eugenol, (C10H12, O2) is dissolved in 12.0 grams of benzene. (Kb benzene= 2.53OC/m; Tb(solvent) = 80.100C)  DEEPEN Let us further deepen our understanding by solving the following problems: The answer for the first is given to you which serves as your guide. Good luck ! No correct solution no credit !!! Activity 2: 1) Calculate the boiling point of a solution containing 12g glucose C6H12O6 dissolved in 200g of water H2O. (0.5120C – Kb of H2O) Answer: 100.1710C 2) Calculate the normal freezing point of a 0.7439 mol aqueous solution of C12H22O11 that has a density of 1.35 g/ml. (C12H22O11 is a non- volatile non-dissociating solute.) The molal freezing point depression constant of water is 1.86OC kg/mole. 3) When 0.279 g of a molecular compound, benzoic acid, was dissolved in 43.0 g of benzene, the freezing point of the solution was lowered to 5.15 °C. What is the molecular weight of the benzoic acid? (Note: Kf for benzene = 5.120C/m; Freezing point of benzene= 5.50C)  GAUGE CONGRATULATIONS! You are now ready to proceed to assess your grasped concepts. Direction: Solve what is asked in each given question. Enclose your answers in a box 15 points per problem ( 3 pts given; 2 points unknown ; 10 points correct solution) 1) When 2.25 grams of an unknown non-electrolyte dissolves in 12.0 grams of water, the solution freezes at −3.16 °C. What is the molecular weight of the unknown compound? Kf for water = 1.86 °C/ m. 2) What is the boiling point elevation when 165 g of lactic acid (C6H10O5) is dissolved in 668 g of cyclohexane (C6H12)? The boiling point constant for cyclohexane is 2.79 °C/m. 3) What is the freezing point of a solution prepared by adding 150. g trichothecin (C19H24O5) to 0.867 kg of benzene? The freezing point of pure benzene is 5.5 °C. The freezing point constant for benzene is 5.12 °C/m. 4) A solution was prepared by dissolving some acetamide, CH2CONH2, in 65.0 g of pure water. The boiling point of the solution at 1 atm was 100.208 °C. How much acetamide, in grams, was dissolved to yield the solution. (0.5120C – Kb of H2O) Jumpstart: Increasing Boiling Point 1m C6H12O6, 1m NaCl, 1m CaCl2, 1m AlCl3 Increasing Freezing Point 1m AlCl3, 1m CaCl2, 1m NaCl, 1m C6H12O6 Explore: 1) 102.540C 2) 80.3530C Deepen: 1) 100.1710C 2) -1.27410C 3) 94.9 g/mol Gauge: 1) 110.29 g/mol 2) 4.250C 3) 2.830C 4) 1.5312g Key Answers References: A. Printed Materials Ayson, Marissa F., De Borja, Rebecca S.(2016). General Chemistry 2. Textbook. Araneta Ave. Quezon City, Philippines: Vibal. Publishing House, Inc. Moore, John W. (2019). K-12 General Chemistry 2. Manila, Philippines:. Rex Bookstore, Inc. B. Online and Electronic Source Dissociation in Chemistry. (n.d) Retrieved January 07, 2021 from. https://www.britannica.com/science/dissociation Boiling Points for common Liquids and Gases. (n.d) Retrieved January. 07, 2021 from www.engineeringtoolbox.com/boiling-points-. fluids-gases-d_155.html Difference Between Vapor Pressure and Boiling Point. (n.d) Retrieved. January 07, 2021 from https://pediaa.com/difference-between-. vapor-pressure-and-boiling-point/#:~:text=The%20boiling%20. point%20is%20the%20temperature%20at%20which,whereas. %20boiling%20point%20is%20a%20measurement%20of%20. temperature. Freezing Science: The Role of Salt in Making Ice Cream. (n.d) Retrieved. January 07, 2021 from www.thekitchn.com/freezing-science-. the-role-of-s-124357