Week 2.pdf

GENERAL CHEMISTRY 2 WEEKs 2 & 3 EDGAR A. RIVERA   Interpret the phase diagram of water and carbon dioxide  Determine and explain the heating and cooling curve of a substance  Use different ways of expressing concentration of solutions: percent by mass,...

GENERAL CHEMISTRY 2 WEEKs 2 & 3 EDGAR A. RIVERA   Interpret the phase diagram of water and carbon dioxide  Determine and explain the heating and cooling curve of a substance  Use different ways of expressing concentration of solutions: percent by mass, mole fraction, molarity, molality, percent by volume, percent by mass, ppm  Perform stoichiometric calculations for reactions in solution  Describe the effect of concentration on the colligative properties of solutions MELCs  Differentiate the colligative properties of nonelectrolyte solutions and of electrolyte solutions  Calculate boiling point elevation and freezing point depression from the concentration of a solute in a solution  Calculate molar mass from colligative property data  Describe laboratory procedures in determining concentration of solutions   MELC: Interpret the phase diagram of water and carbon dioxide (STEM_GC11 IMFIIIa-c-107) LESSON 1: Phase  OBJECTIVES: 1. describe the features of a phase diagram diagram of 2. interpret the phase diagram of water and water and carbon dioxide and changes it undergoes carbon 3. appreciate the importance of crystalline and dioxide amorphous solids in daily life   How can this effect be achieved using CO2 or dry ice? What are your thoughts? Because carbon dioxide cannot exist as a liquid at atmospheric Let’s pressure, the dry investigate! ice sublimes and instantly produces a gas, condensing water vapor, and creating a thick white fog.   Phase changes are transformations of matter from one physical state to another. They occur when energy (usually in the form of heat) is added or removed from a substance. Can you recall the types of phase changes? Fill-out the diagram with the correct types of phase changes that you know. Sublimation TRY THIS! Melting Vaporization Freezing Condensation Deposition   A phase diagram is a graphical representation of the physical states of a substance under different conditions of temperature and pressure. PHASE DIGRAM   Direction: Follow the steps indicated to construct a phase diagram using the materials needed. Analyze the different features of the phase diagram based on your constructed output and describe the Activity No. _ different features of a phase diagram and answer the questions. Diagram  Construction  Materials: 3 different-colored crayons, green, blue and red pens/ pencils, a pen, ruler   The Three Areas The three areas are marked solid, liquid, and vapor. Under a set of conditions in the diagram, a substance can exist in a solid, liquid, or vapor (gas) phase. The labels on the graph represent the stable states of a system in equilibrium. What are the features of a phase diagram? Phase diagram with three sets of conditions   Three Lines (Curves) 1. The green line divides the solid and liquid phases, and represents melting (solid to liquid) andfreezing (liquid to solid) points. 2. The blue line divides the liquid and gas phases, and represents What are the vaporization (liquid to gas) andcondensation (gas to liquid) points. features of a phase diagram? The freezing (or melting) curve   Vaporization (or condensation) curve – the curve on a phase diagram which represents the transition between gaseous and liquid states.  It shows the effect of pressure on the boiling point of the liquid.  Anywhere along this line, there will be equilibrium between the liquid and the vapor. What are the features of a 3. The red line divides phase the solid and gas diagram? phases, and represents sublimation (solid to gas) and deposition (gas to solid) points. The vaporization or condensation curve   Sublimation (or deposition) curve – the curve on a phase diagram which represents the transition between gaseous and solid states.  It represents the effect of increased temperature on a solid at a very low constant pressure, lower than the triple point. What are the features of a phase diagram? Sublimation or deposition curve.   Two Important Points 1. The triple point The triple point is the combination of pressure and temperature at which all three phases of matter are at equilibrium. It is the point on a phase diagram at which the three states of matter coexist. What are the The lines that represent the conditions of solid-liquid, liquid-vapor, features of a and solid-vapor equilibrium meet at the triple point phase diagram? Temperature and pressure values at the t ripplepoint   Two Important Points 2. The critical point  The critical point terminates the liquid/gas phase line. It is the set of temperature and pressure on a phase diagram where the liquid and gaseous phases of a substance merge together into a single phase. Beyond the temperature of the critical point, the merged What are the single phase is known as a supercritical fluid. features of a  phase  The temperature and pressure corresponding to this are known as diagram?  the critical temperature and critical pressure.  If the pressure on a gas (vapor) is increased at a temperature lower than the critical temperature, the liquid- vapor equilibrium line will eventually be crossed and the vapor will condense to give a liquid.   Two Important Points 2. The critical point What are the features of a phase Temperature and diagram? pressure values at the critical point   The normal melting and boiling points are those when the pressure is 1 atmosphere. How is the normal melting and boiling points determined in Locating the normal melting point and a phase normal boiling point diagram?  The normal melting and boiling points of water are found in exactly the same way as we have already discussed - by determining where the 1 atm pressure line crosses the solid-liquid, and then the liquid-vapor equilibrium lines. The normal melting point of water is 273 K (0 oC), and its normal boiling point is 373 K (100 oC). How does the phase diagram of Phase water look diagram for like? H2O  The Phase Diagram for Carbon Dioxide How does At 1 atm pressure, carbon the phase dioxide will sublime at a temperature of 197.5 K (- diagram for 75.5 °C). This is the reason carbon why solid carbon dioxide is often known as "dry ice." dioxide look There is no liquid carbon like? dioxide under normal conditions - only the solid or the vapor. Phase diagrams for CO2   Interpreting a Phase Diagram Direction: Refer to the following phase diagram of a certain substance to answer the following questions. 2. At what pressure 4. 3. What 1. Iniswhat What phase will phase(s) the normal ismelting and temperature point ofthe exist atsubstance the1substance? conditions atm and will at all70 °C? 50 °C and 1 atm three phases of the pressure? Try This! substance be present? liquid ≈ 0.5 and°Cvapor atm liquid ≈ 32 (gas) and ≈28 °C Performance Task No. __: Interpreting Phase Diagrams of Water and Based from the phase diagrams of water and carbon dioxide, answer the Carbon Dioxide following questions and justify your answers: 1. You have ice at 263 K (-10.0 oC) and 1.0 atm. What could you do to make the ice sublime? 2. A sample of dry ice (solid CO2) is cooled to 173 K (-100.0 oC), and is set on a table at room temperature (298 K; 25 oC). At what temperature is the rate of sublimation and deposition the same (assume that pressure is held constant at 1 atm)? END LESSON 1   How does a change in energy affect phase changes? When a substance is heated, the added energy is used by the substance in either of two ways: The added heat increases the kinetic energy of the particles and the particles a. LESSON 2: move faster. The increase in kinetic energy is accompanied by an increase in Heating and temperature. The added heat is used to break attractive forces between particles. There is b. cooling no observed increase in temperature when this happens. Often a change in curve of a the physical appearance of the substance is observed, such as a phase change. substance Conversely, the removal or release of heat results in two ways: a. A decrease in kinetic energy of the particles. The motion of the particles slow down. A decrease in temperature is observed. b. Forces of attraction are formed, and a phase change may occur. No change in temperature is observed.  Heating curve is the change in temperature of a substance as it is being heated can be shown in a graph. The heating curve is a plot of temperature and heat added to the substance. Often, time is used instead of heat added in the abscissa, because it is assumed that heat is uniformly added per unit time. HEATING CURVE OF A SUBSTANCE Heating curve COOLING CURVE OF A SUBSTANCE Cooling curve (Image Source: http://www.ausetute.com.au/images/coolcurv.gif)   Answer the assessment by clicking the link below. ASSESSMENT  PHASE DIAGRAM OF WATER AND CARBON DIOXIDE (google.com)   MELC: Use different ways of expressing concentration of solutions: percent by mass, mole fraction, molarity, molality, percent by volume, percent by mass, ppm OBJECTIVES: LESSON 3:  Identify the different ways of expressing the concentration of a concentration solution. of solutions  Carry out stoichiometric calculations involving reactions in solution.  Recognize the importance of understanding the different ways of expressing solution concentration.  1. Percent by Mass (m/m). 2. Percent by Volume(v/v). Ways of 3. Percent mass by volume(m/v). Expressing 4. Parts Per Million, ppm. Solution Concentration 5. Molarity. 6. Molality. 7. Mole Fraction.   Percent by Mass (m/m). It is referred as the mass of solute, in grams(g), present in 100 g of solution. PERCENT BY MASS (M/M)   A sample of 0.446 g sodium chloride (NaCl) is dissolved in 27.3 g water (H2O). What is the percent by mass of NaCl? SAMPLE PROBLEM   Percent by Volume(v/v). It is referred as the volume of solute present in 100 mL of solution. PERCENT BY VOLUME (V/V)   A 1.75-L bottle wine contains 560 mL of ethanol. What is the v/v percent concentration of ethanol? SAMPLE PROBLEM   Percent mass by volume(m/v). It is referred as the mass of solute, in g, present in 100 mL of solution. PERCENT MASS BY VOLUME (M/V)   If 30.0 g calcium chloride, CaCl2, is present in 0.5000 L of aqueous solution, what is its concentration in terms of mass/volume percent? SAMPLE PROBLEM   Parts Per Million, ppm. This unit of concentration is referred as the parts of a component per million parts (106 or 1,000,000) of the solution and is usually used if the quantity of solute present in a solution is very PARTS PER small. MILLION, PPM   If there is 1.102 mg of silver (Ag) in 696 g of solution, what is the concentration of Ag in ppm? SAMPLE PROBLEM   Molarity. It is another unit of concentration defined as moles of solute per liter of solution. MOLARITY   What is the molarity of a solution containing 37.4 g of NaCl in 6.8 liters solution? SAMPLE PROBLEM   Molality. It is the number of moles of solute dissolved in 1 kilogram of solvent. MOLALITY   Calculate the molality of a sulfuric acid solution containing 48.8 g of sulfuric acid in 396 g of water. SAMPLE PROBLEM   Mole Fraction. It is a unitless quantity that expresses the ratio of the number of moles of one component to the total number of moles of the solution. MOLE FRACTION   A solution is prepared by adding 400.8 g ethanol, C2H5OH to 287.8 g H2O. Calculate the mole fractions of these two components. (Molar masses: C2H5OH = 46.07 g/mol; H2O =18.02 g/mol) SAMPLE PROBLEM 0.6396   Direction: Solve the following problems and show your solutions in 1 whole sheet of paper. EXERCISES   Direction: Answer the assessment by clicking the link below. CONCENTRATION OF SOLUTION ASSESSMENT (google.com) END LESSON 3   MELC: Perform stoichiometric calculations for reactions in solution (STEM_GC11PPIIId-f-112 ) LESSON 4: OBJECTIVES: Stoichiometric 1. Define stoichiometry calculations 2. Apply the concept of mole to obtain the amount of a given substance. for reactions 3. Solve problems for quantitative relationships in balanced in solution chemical equations using the mole method.   Stoichiometry is the study of the quantitative relationships or ratios between two or more substances undergoing a physical change or chemical change (chemical reaction).  The word derives from the Greek SOLUTION words: stoicheion (meaning "element") and metron (meaning "to measure"). Most often, STOICHIOMETRY stoichiometry calculations deal with the mass or volumes of products and reactants.  In stoichiometric calculation, solution concentration can be used as a conversion factor to determine the amount of product that can be produced in a chemical reaction.   Determine the mass, in grams, of solid Mg(OH)2 that can be produced if 90.0 mL of a 1.26 M Mg(NO3)2 solution completely reacts with excess NaOH. (Atomic mass: Mg = 24.31 g/mol; O = 16.00 g/mol; N = 14.01 g/mol; Na = 22.99 g/mol; H = 1.008 g/mol) Sample Given: 90.0 mL of 1.26 M Mg(NO3)2 Problem: mass of Mg(OH)2 = ? Solution Stoichiometry 1. Write the balanced chemical reaction 2NaOH(aq) + Mg(NO3)2(s) Mg(OH)2(s) +2NaNO3(aq)   2. calculate moles of Mg(NO3)2. 90 mL x 1L 1.26 mol Mg(NO3)2 1000 mL 1L Sample = 0.113 mol Mg(NO3)2 Problem 1: 3. calculate the mass of Mg(OH)2 Solution Stoichiometry 1 mol Mg(OH)2 0.113 mol Mg(NO3)2 1 mol Mg(NO3)2 Molar mass 58.33 g Mg(OH)2 1 mol Mg(OH)2 Stoichiometric ratio   Calculate the mass, in grams, of 6.00% H2O2 solution is needed to produce 132.6 g of O2 (g). Sample 1. Write the balanced chemical equation Problem 2: 2H2O2(aq) 2H2O(l) + O2(g) Solution Stoichiometry 2. Calculate moles O2.  3. Calculate mass of H2O2 solution. Sample Problem 2: Solution Stoichiometry   Directions: Solve the following problems and show your solutions in your notebook. EXERCISES   Direction: Answer the assessment by clicking the link below. ASSESSMENT  STOICHIOMETRIC CALCULATIONS FOR REACTIONS IN SOLUTIONS (google.com) END LESSON 4   MELCs:  Describe the effect of concentration on the colligative properties of solutions (STEM_GC11PPIIId-f-115)  Differentiate the colligative properties of non-electrolyte and LESSON 5: of electrolyte solutions (STEM_GC11PPIIId-f116) OBJECTIVES: Colligative  define colligative properties. properties of  determine the different colligative properties of solutions; and solutions  describe how the concentration of solutions affects the colligative properties of solutions  appreciate the importance of colligative properties in everyday life . THINK ABOUT THIS!   COLLIGATIVE PROPERTIES = are the properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. VAPOR PRESSURE LOWERING COLLIGATIVE PROPERTIES BOILING POINT ELEVATION FREEZING POINT DEPRESSION OSMOTIC PRESSURE   Nonvolatile - (does not have a measurable vapor pressure), the vapor pressure of its solution is always less than that of the pure solvent.  Volatile - substance that evaporates readily. VAPOR PRESSURE LOWERING   Question #1: Based from the discussion and illustration, why is it that when you try to boil an egg in an open container/pot, and leave awhile, upon returning you’ve noticed the water decreases? What about the movement of water molecules? What if you decided to put a lid on the Think About container/pot, what will be the movement of the This water molecules? Question #2: What is the relationship between the volatility of a substance to the vapor pressure? How about a nonvolatile substance to its vapor pressure?  Pressure cooker operates on the principle that an increase in the pressure on the surface of a liquid will increase the boiling point of the liquid. Example of Vapor Pressure  A coolant/antifreeze circulates through the engine, maintaining the correct working temperature of different components. The main component of a coolant/antifreeze is glycol. Example of Vapor Pressure If we don’t apply coolant/antifreeze on the car radiator there would have an abnormal temperature, automatic engine cutoff and worst may bring damage to the engine parts.  if the liquid is water containing salt, as is true of Great Salt Lake, its vapor pressure will be lower than a similar body of fresh water at the same temperature. The molecular activity of water is being reduced when its salinity increases because molecules of dissolved solids interfere with the motion of water molecules. Example of Vapor Pressure Great Salt Lake in Northern Utah, USA  Of these two solvents, alcohol has a greater tendency to have its molecules at the liquid surface escape into the gas phase. the vapor pressure for alcohol is greater than that of water for a given temperature. Example of Vapor Pressure   states that the vapor pressure of a solvent RAOULT’S above a solution is equal to the vapor pressure of the pure solvent at the same LAW temperature scaled by the mole fraction of the solvent present   The boiling point is the temperature at which the vapor pressure of a liquid is equal to the atmospheric pressure. BOILING  If a liquid has a high vapor pressure it means that POINT the molecules evaporate faster and it takes a shorter time to equalize the vapor pressure of the ELEVATION liquid and the atmospheric pressure.  The boiling point elevation is the difference in temperature between the boiling point of the pure solvent and that of the solution.  The ethylene glycol helps prevent the water in the car’s radiator from freezing through freezing point depression, but it can also elevates the boiling point of the fluid as well. By raising the boiling point, it helps to protect against over heating. Example of Boiling Point Elevation  Adding salt to water will increase the water’s boiling point or temperature. Boiling point elevation is not making the water to boil faster instead adding salt will take slightly longer to boil, since its boiling point has now been elevated. Example of Boiling Point Elevation  Sugar is from the sugarcane crop that has been harvested and the cane juice was been extracted. The boiling point elevation is one way of monitoring the level of saturation of the solution which is an important consideration for Example of crystallization. Boiling Point Elevation THINK ABOUT Why is it that burn caused by THIS! boiling syrup is more severe than a burn caused by boiling water? FREEZING  Freezing point depression is colligative property of solution, which describes the ability POINT of a dissolved solute to lower the freezing point DEPRESSION of its solution.  They added salt on icy roadways. The salt that is being added on the roads will result for the water to not freeze on its normal freezing point (0°C) instead it lowers to as low as -9°C. The reason of this application is that salt causes ice to melt because salt has a lower freezing point than pure water. Ice in contact with salty water therefore melts, creating more liquid water, which dissolves more salt, thereby causing more ice to melt, and so on. Example of The higher the concentration of dissolved salt, the lower its overall freezing point Freezing Point Depression  Another example is the de-icing of planes. A number of solutions are used but commonly a solution such as ethylene glycol, or a less toxic mono propylene glycol, is used to deice an aircraft. The aircrafts are sprayed with the solution when the temperature is predicted to drop below the freezing point. Example of Freezing Point Depression   Have this thought comes into your mind why is it hot countries put salt on ice in the ice cream makers to keep the ice from melting, while in cold countries put salt on ice to melt it? THINK ABOUT Ans: This is because adding salt to an ice water THIS! mixture lowers the freezing point (or melting point) of the equilibrium.  Osmotic Pressure - the pressure required to prevent osmosis. Osmosis is the diffusion of water molecules across a selectively permeable membrane from an area of higher concentration to an area of lower concentration that OSMOTIC allows the passage of solvent molecules through a PRESSURE porous membrane from a dilute solution to a more concentrated one. Low Highly concentrated concentrated solution solution OSMOTIC PRESSURE The hypertonic solution is the opposite of a hypotonic solution, where there is more solute outside the cell than inside it. Isotonic solution has the same concentration of solutes both inside and outside of the cell. Hypotonic solution there is a higher concentration of solutes inside the cell than outside the cell.  Have you observed yourself when you submerged in a water for a certain longer time? EXAMPLE OF OSMOTIC PRESSURE This is due to the water molecules which naturally flow through a semipermemable mebrane from a low concentration area to a high concentration area through a process called osmosis. The osmosis occurs and the water flows into the upper skin cells, which then take up the water.  The concentration of the solution inside the cucumber is less than the concentration of the brine solution, so water migrates through the cell walls into the brine causing the cucumber to shrink. EXAMPLE OF OSMOTIC PRESSURE  A fish that used to lives in salt water will have somewhat salty water inside itself when put it on the freshwater, and the freshwater will, through osmosis, enter the fish, causing its cells to swell and eventually dies. EXAMPLE OF OSMOTIC PRESSURE  Home preservation like jam and jelly is a good example of osmotic pressure. The bacterial cell is in a hypertonic (high-concentration) sugar solution, the intracellular water tends to move out of the bacterial cell to the more concentrated solution by osmosis. EXAMPLE OF This process known as crenation, causes the cell to shrink, and eventually cease functioning. OSMOTIC PRESSURE   MELC: Calculate boiling point elevation and freezing point depression from the concentration of a solute in a solution OBJECTIVES: LESSON 7:  discuss the applications of colligative properties and Boiling point laboratory procedure in determining concentration of solutions elevation  determine the boiling point elevation and freezing point and freezing depression from the concentration of a solute in a solution, and calculate molar mass from colligative point property data. depression  promote the importance of colligative properties of solutions. PICTURE THIS OUT!   What is the difference between the boiling point of a solution with that of a pure solvent? BOILING POINT ELEVATION The boiling point elevation is the difference in temperature between the boiling point of the pure solvent and that of the solution.  The boiling point elevation of a solution, ΔTb, is directly proportional to molality, that is, Where Kb is the molal boiling point elevation constant of the solvent and m is the molal concentration of the solution (mol/kg) BOILING POINT ELEVATION Table 1: Boiling Point Elevation Constant (Kb) and Freezing Point Depression Constant (Kf) for some solvents.   What will be the boiling point of a solution prepared by dissolving 10 g of NaCl in 25 g of water? Given: Sol’n: SAMPLE Mass of NaCl = 10 g PROBLEM Mass of H2O = 25 g Kb H2O = 0.51 °C/molal nsolute m= kgsolvent ΔTb = ? 0.17 mol m= = 6.8 m 0.025 kg ΔTb = (0.51 °C/m) (6.8 m) ΔTb = 3.47 °C + 100 °C ΔTb = 103.47 °C   The freezing point of a substance is the temperature at which the solid and liquid forms can coexist indefinitely, at equilibrium. FREEZING POINT DEPRESSION   Like the boiling point elevation, the freezing point of a solution is directly proportional to the molal concentration of the solution, that is, FREEZING POINT where : DEPRESSION ΔTf refers to the freezing point lowering, Kf, the freezing point depression constant m, the molality of the solution.   What will be the boiling point and freezing point of a solution prepared by dissolving 10 g of NaCl in 25 g of water? SAMPLE PROBLEM   A solution prepared from 0.3 g of unknown nonvolatile solute and 30 g of CCl4 has a TRY THIS! boiling point of 0.392 °C higher than that of pure CCl4. What is the molar mass of the solute? LESSON 8:  MELC: Calculate molar mass from colligative property Calculation data of molar OBJECTIVES: mass from 1. Describe molar mass colligative 2. Solve problems involving calculating molar mass from colligative property data property 3. Open-mindedness to the new information data  Sample Problem: A prepared solution consists of 1.25 g of methyl Determining salicylate (C8H8O3) in 99.0 g of Molar Mass benzene (C6H6) has a boiling point from Boiling of 80.31 °C. Determine the molar Point Elevation mass of methyl salicylate in the solution. STRATEGY (sequence of conversion)  Sample Problem: A 7.85 g unknown sample of a compound is dissolved Determining in 301 g of benzene. The freezing Molar Mass point of the solution is 1.05 °C below from Freezing that of pure benzene (C6H6). Point Determine the molar mass of the Depression unknown compound present in the solution. STRATEGY (sequence of conversion)  Sample Problem: A solution is prepared by dissolving 35.0 g of an Determining unknown compound in enough Molar Mass water to make up 1 L in volume. from Osmotic Determine the of the molar mass of Pressure the unknown solute if the osmotic pressure of the resulting solution is 10.0 mmHg at 25 °C STRATEGY (sequence of conversion)  Directions: Summarize the steps in determining the molar mass of an unknown solute from the different colligative property below by supplying the missing terms. W RA P –U P  . Directions: Perform the needed calculations for the following problems and supply the missing answer on the table below. Show your complete calculations in a clean sheet of paper EXERCISE:   Directions: Perform the needed calculations for the following problems and supply the missing answer on the table below. Show your complete calculations in a clean sheet of paper EXECISE:  In a clean sheet of paper, determine the importance of identifying the molar mass of a VALUING particular unknown solute in a solution and think of any practical application of this concept on our daily life  Direction: Answer the assessment by clicking the link below. COLLIGATIVE PROPERTIES OF SOLUTION ASSESSMENT (google.com) LESSON 9: Laboratory procedures  MELC: Describe laboratory procedures in determining in concentration of solutions determining concentration of solutions LESSON 10: Laboratory procedures  MELC: Describe laboratory procedures in determining in concentration of solutions determining concentration of solutions Steps involved in the preparation of a standard solution.  A useful relationship in the preparation of dilute solutions is expressed by the following:   Dilution of Solution  Given the formula below, try to find out on how you will prepare 1.0 L (1000cm3) of 0.1 M acetic acid (CH3COOH).The concentration of commercial concentrated CH3COOH is 17 M. Example GEN. CHEMISTRY 2: Laboratory Procedures in Determining Concentration of Solutions  Direction: Identify the terms that correspond to the definitions which are given on the second column. Use the word list as guide. RECAP   1. Preparation of a Standard Solution TRY THIS!   2. Dilution of Solution Given the formula below, try to find out on how you will prepare 1.0 L (1000cm3 ) of 0.1 M acetic acid (CH3COOH).The concentration of commercial concentrated CH3COOH is 17 M TRY THIS! Preparation of a standard solution  In diluting an acid, always WHAT DO add acid to water. Never do YOU THINK? the reverse. Why do you think so? Relationship in the preparation of dilute solutions   Given the formula below, try to find out on how you will prepare 1.0 L (1000cm3 ) of 0.1 M acetic acid (CH3COOH).The concentration of commercial concentrated CH3COOH is 17 M. Dilution of Solution Therefore, you need to measure 5.9 cm3 of concentrated CH3COOH and add to 994.1 cm3 of distilled water to make 1000 cm3 of solution.  What volume of 0.25 M HCl solution must be diluted to prepare 1.00 L of APPLY WHAT 0.040 M HCl? YOU LEARNED! How will you prepare 250 cm3 of a 1 M NaOH solution using NaOH pellets?   In the winemaking industry, laboratory procedures are employed in determining the right concentration of wine for safety purposes. For example, to analyze the alcohol content of a certain wine, a chemist needs 1.00L of an aqueous 0.200 M K2Cr2O7 (potassium dichromate) solution. How much solid K2Cr2O7 must be weighed out to make this solution? ELABORATE   Direction: Design a procedure on how you will prepare 1.25 M table sugar (C12H22O11) solution and 1.25 M table salt (NaCl) solution diluted to 1 L water. Perform three trials each set-up and get the average. Write/ encode your output in short bond paper with the following parts: Title GROUP ____________________ PERFORMAN I. Objectives: II. Materials: CE TASK III. Procedure (including pictures/diagrams): IV.Data: V. Generalization: (Note: Perform this activity on a separate table/ area at home to ensure accuracy and precision of data. Be careful on handling the solution to avoid spillage.)   Direction: Answer the assessment by clicking the link below. LABORATORY PROCEDURES IN ASSESSMENT DETERMINING CONCENTRATION OF SOLUTIONS (google.com)