Expression of Concentration PDF

Learning Guide Module Subject Code Chem 1 General Inorganic Chemistry Module Code 9.0 Solutions Lesson Code 9.3 Expression of Concentration Time Limit 30 min Components Tasks...

Learning Guide Module Subject Code Chem 1 General Inorganic Chemistry Module Code 9.0 Solutions Lesson Code 9.3 Expression of Concentration Time Limit 30 min Components Tasks ATa ATAb Target At the end of the chapter, the students must be able to express the concentration of solutions in terms of molarity, molality, mole fraction, 1 min percent concentration, ppm, ppb, and proof. WHAT ARE YOU MADE OF? The Parable of Potatoes, Eggs, and Coffee Beans It was one of the quarantine nights, when a child complained to his mother that he was facing life in a pitiable state of distress. The days were cruel and so was his mind. He was no longer happy of what he was doing. Struggling to meet expectations, he burned himself out. Unable to focus clearly, the pandemic was the one to blame for his anxious dispositions and loss of productivity. His mother, a chef de partie, took him to the kitchen. Three cauldrons were filled water and were subjected to high fire. She placed the potatoes in one cauldron, eggs in the second, and ground coffee beans on the third one. The water in each cauldron started to boil. The child made a low prolonged sound of impatience, wondering what his mother was doing. After some time, she took the potatoes and eggs out Hook of the cauldron and placed them in a bowl. The coffee was transferred into a cup. 3 min Turning to him, she asked the child. “What are these?” “Potatoes, eggs and coffee,” he annoyingly made a response. Fig. 1. The Potato, the Egg & the Coffee Beans [graphic illustration]. (2016). Retrieved December 24, 2020 from ScoopWhoop. “Observe closely”, she said, “and try touching the potatoes.” The child did and noticed that the potatoes were soft. Chem 1 Expression of Concentration Page 1 of 12  She then asked him to get an egg and have it out of its shell. After pulling off the shell, he observed an intact state of a hard-boiled egg. Finally, she offered the coffee. Its savory smell brought a smile to his face. The mother explained that the potatoes, the eggs and coffee beans had each encountered the same adversity, the same stressor – the boiling water. However, each one of them reacted in different ways. At first, the potato was hard. It appeared to have a great physical power and tend to yield in determination. But in boiling water, it became weak and soft. The egg must be handled carefully for it was fragile. When placed in a boiling water, the insides of the egg became hard. However, the ground coffee beans were worthy of note. After being exposed to the boiling water, the water changed and something new was created. “What are you made of?” she asked her son. “When the boiling water tossed you in different directions, how will you respond? Are you a potato, an egg, or a coffee bean?” The moral of the story: Life throws countless concerns, troubles, and triggers to us. The only thing that will hold of great value will be our ways of reacting to the adversity on hand. Learn how you can adapt and make the best out of these experiences for the same water can soften the potatoes, harden the eggs, and make something new out of the ground coffee beans. I have my take on this as a chemist. Having your boiling water as the solvent and your ground coffee beans as the solute, it is safe to say that you were able to create a solution out of the adversity. You were able to produce an aroma that brought smiles to faces. You were able to make yourself suitable to an uncomfortable situation and regarded it as not bothersome. Be a coffee bean! One thing, though. I am quite curious about the concentration of the coffee when the ground coffee beans were mixed with water. Are you? Come on, let’s discover how we can express our concentration units! A mixture, different from a chemical compound, is composed of two or more components. The relative amounts of these substances Ignite involved in a solution must be properly specified. 17 QUALITATIVE min There are several ways to express the amount of solute present in a solution. Concentration refers to the measure of the amount of solute (substance that is being dissolved) in a given amount of solvent Chem 1 Expression of Concentration Page 2 of 12  (substance that dissolves). A concentrated solution contains a somewhat large amount of dissolved solute while a dilute solution has less. Fig. 2. Solutions of a red dye in water from the most dilute (on the left) to the most concentrated (on the right). [photograph]. (2020). Retrieved December 24, 2020 from Chem LibreTexts. Solutions are also characterized by their ability to dissolve a solute. Fig. 3. Types of Saturation. [graphic illustration]. (2019). Retrieved December 24, 2020 from Pinterest. An unsaturated solution contains solute that is lesser than its capacity to dissolve. A saturated solution, at a specific temperature, contains the greatest possible amount of a solute that will dissolve in a given solvent. Once the maximum amount of solute is reached, it will remain undissolved in the solution A supersaturated solution comprises more solute dissolved than that in a saturated solution. This is usually achieved when heat is applied. This type of saturation is unstable and tends to undergo crystallization and precipitation. Chem 1 Expression of Concentration Page 3 of 12  Fig. 4. Sodium acetate crystals rapidly result when a small seed crystal is added. [graphic illustration]. (2019). Retrieved December 24, 2020 from Chemistry, 10th Edition. QUANTITATIVE In performing calculations, we need to be more specific with the solution’s contents rather than by just using the terms concentrated or dilute. 1. Percent Composition One of the many ways to present the concentration of a solution is by the percent of the solution that is constituted of a solute dissolved in a solvent. This percentage can be described in one of these three ways: a. Percent by Mass This refers to the ratio of the mass of a solute to the sum of the masses of the solute and the solvent or the mass of the solution, multiplied by 100 percent. 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑏𝑦 𝑚𝑎𝑠𝑠 = 𝑥 100% 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 + 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑏𝑦 𝑚𝑎𝑠𝑠 = 𝑥 100% 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 Example: Potassium chloride is a crystalline salt that is found in natural waters and is used as fertilizer. You have a sample of 0.892 g of potassium chloride (KCl). You decided to dissolve it in 54.6 g of water. Find the percent by mass of KCl in the solution. Solutions: The mass of the solute dissolved in the solvent is given. You can easily calculate the mass percent using this equation: 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑏𝑦 𝑚𝑎𝑠𝑠 = 𝑥 100% 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 Substituting the values, we write: 0.892 𝑔 𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑏𝑦 𝑚𝑎𝑠𝑠 = 𝑥 100% 0.892 𝑔 + 54.6 𝑔 𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑏𝑦 𝑚𝑎𝑠𝑠 = 1.61 % b. Percent by Volume It is the ratio of the volume of a solute to the sum of the volume of the solute and the solvent or the volume of the solution, multiplied by 100 percent. Chem 1 Expression of Concentration Page 4 of 12  𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑏𝑦 𝑣𝑜𝑙𝑢𝑚𝑒 = 𝑥 100% 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 + 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑏𝑦 𝑣𝑜𝑙𝑢𝑚𝑒 = 𝑥 100% 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 Example: It was Christmas Eve and you found a 0.750-L bottle of Italian chianti under the basement. Its label says “11.5% alcohol by volume.” Find the amount of alcohol, in liters, that the wine contains. Solutions: The volume % of the wine was given: 11.5%. You can see this as 11.5 parts alcohol to 100. parts of chianti. The total volume of the solution in the bottle was 0.750 L. We can easily find the amount of alcohol using this equation: 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑏𝑦 𝑣𝑜𝑙𝑢𝑚𝑒 = 𝑥 100% 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 Substituting and transposing the values, we write: 11.5 𝐿 𝑎𝑙𝑐𝑜ℎ𝑜𝑙 𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑏𝑦 𝑣𝑜𝑙𝑢𝑚𝑒 =.750 𝐿 𝑐ℎ𝑖𝑎𝑛𝑡𝑖 𝑥 100. 𝐿 𝑐ℎ𝑖𝑎𝑛𝑡𝑖 𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑏𝑦 𝑣𝑜𝑙𝑢𝑚𝑒 = 0.0863 𝐿 𝑎𝑙𝑐𝑜ℎ𝑜𝑙 c. Percent by Mass/Volume It is also used in some occasions and is computed in the same manner as to the two percentages presented earlier. This is calculated by simply dividing the mass of the solute by the volume of the solution, multiplied by 100 percent. 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑏𝑦 𝑚𝑎𝑠𝑠/𝑣𝑜𝑙𝑢𝑚𝑒 = 𝑥 100% 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 Example: A brine solution was prepared using 10 grams of sodium chloride in enough water to make a 150mL solution. Find the mass-volume percentage. Solutions: 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑏𝑦 𝑚𝑎𝑠𝑠/𝑣𝑜𝑙𝑢𝑚𝑒 = 𝑥 100% 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 10 𝑔 𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑏𝑦 𝑚𝑎𝑠𝑠/𝑣𝑜𝑙𝑢𝑚𝑒 = 𝑥 100% 150 𝑚𝐿 𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑏𝑦 𝑚𝑎𝑠𝑠/𝑣𝑜𝑙𝑢𝑚𝑒 = 6.7% Chem 1 Expression of Concentration Page 5 of 12  2. Parts per Million (ppm), Parts per Billion (ppb) These concentration units are being used when we are dealing with extremely dilute solutions. A part per million (ppm) is one part of solute per million part of solution. The first few equations can still be applied in computing ppm. 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑝𝑝𝑚 (𝑚/𝑚) = 𝑥 106 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑝𝑝𝑚 (𝑣/𝑣) = 𝑥 106 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑝𝑝𝑚 (𝑚/𝑣) = 𝑥 106 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 A part per billion (ppb) is one part of solute per billion part of solution. The factor to be used will be 109 rather than 106. 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑝𝑝𝑏 (𝑚/𝑚) = 𝑥 109 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑝𝑝𝑏 (𝑣/𝑣) = 𝑥 109 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑝𝑝𝑏 (𝑚/𝑣) = 𝑥 109 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 Example: A 3.50-gram pill contains 40.5 milligrams of calcium ion. Now, in ppm, find the concentration of Ca2+. Solutions: The mass of calcium ion was given, 40.5 mg. The mass of the pill was 3.50 g. The mass of Ca2+ must be converted first from mg to g for easy cancellation. Now, find the mass ratio of Ca2+ to pill. Then, multiply to the factor of 106 to obtain ppm. 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑝𝑝𝑚 (𝑚/𝑚) = 𝑥 106 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐶𝑎2+ 𝑝𝑝𝑚 (𝑚/𝑚) = 𝑥 106 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑝𝑖𝑙𝑙 1𝑔 40.5 𝑚𝑔 𝐶𝑎2+ 1000 𝑚𝑔 𝑝𝑝𝑚 (𝑚/𝑚) = 𝑥 106 3.50 𝑔 𝑝𝑖𝑙𝑙 𝑝𝑝𝑚 (𝑚/𝑚) = 1.16𝑥104 𝑝𝑝𝑚 𝐶𝑎2+ Chem 1 Expression of Concentration Page 6 of 12  3. Molarity (M) It the amount, in moles, of solute dissolved in 1 L of solution. 𝑎𝑚𝑜𝑢𝑛𝑡 (𝑚𝑜𝑙𝑒) 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑚𝑜𝑙 𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 (𝑀) = = 𝑣𝑜𝑙𝑢𝑚𝑒 (𝐿) 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝐿 Example: What will be the molarity of a solution is prepared 42.23 g of NH4Cl was dissolved in water to come up with a 500.0-mL solution? Solutions: Take note that the mass of ammonium chloride must be first converted into moles. The given volume must be converted to liters as well. 1 𝑚𝑜𝑙 𝑁𝐻4 𝐶𝑙 42.23 𝑔 𝑁𝐻4 𝐶𝑙 𝑥 = 0.7893 𝑚𝑜𝑙 𝑁𝐻4 𝐶𝑙 53.50 𝑔 𝑁𝐻4 𝐶𝑙 𝑎𝑚𝑜𝑢𝑛𝑡 (𝑚𝑜𝑙𝑒)𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 (𝑀) = 𝑣𝑜𝑙𝑢𝑚𝑒 (𝐿)𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 0.7893 𝑚𝑜𝑙 𝑁𝐻4 𝐶𝑙 𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 (𝑀) = 0.5000 𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 (𝑀) = 1.579 𝑀 4. Molality (m) If you can notice, this concentration unit does not involve volume in its ratio. It is the amount, in moles, of solute dissolved in 1 kg of solvent. 𝑎𝑚𝑜𝑢𝑛𝑡 (𝑚𝑜𝑙𝑒) 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑚𝑜𝑙 𝑀𝑜𝑙𝑎𝑙𝑖𝑡𝑦 (𝑚) = = 𝑚𝑎𝑠𝑠 (𝑘𝑔) 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 𝑘𝑔 Example: Calcium chloride, a white salt that tends to undergo gradual melting, is used as a drying and dehumidifying agent to control dust and ice on pathways. If you prepare a solution by dissolving 32.0 grams of CaCl2 in 271 grams of water, what will be its molality? Solutions: The mass of calcium chloride must be converted to amount (in moles). Then it will be divided by the mass of the solvent, converted in kilograms. 1 𝑚𝑜𝑙 𝐶𝑎𝐶𝑙2 𝐴𝑚𝑜𝑢𝑛𝑡 (𝑚𝑜𝑙)𝑜𝑓 𝐶𝑎𝐶𝑙2 = 32.0 𝑔 𝐶𝑎𝐶𝑙2 = 0.288 𝑚𝑜𝑙 𝐶𝑎𝐶𝑙2 110.98 𝑔 𝐶𝑎𝐶𝑙2 Chem 1 Expression of Concentration Page 7 of 12  𝑎𝑚𝑜𝑢𝑛𝑡 (𝑚𝑜𝑙𝑒) 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑀𝑜𝑙𝑎𝑙𝑖𝑡𝑦 (𝑚) = 𝑚𝑎𝑠𝑠 (𝑘𝑔) 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 0.288 𝑚𝑜𝑙 𝐶𝑎𝐶𝑙2 𝑀𝑜𝑙𝑎𝑙𝑖𝑡𝑦 (𝑚) = 1 𝑘𝑔 271 𝑔 𝑥 1000 𝑔 𝑀𝑜𝑙𝑎𝑙𝑖𝑡𝑦 (𝑚) = 1.06 𝑚 𝐶𝑎𝐶𝑙2 5. Mole Fraction It describes the ratio of the number of moles of solute to the total number of moles (solvent plus solute). 𝑎𝑚𝑜𝑢𝑛𝑡 (𝑚𝑜𝑙𝑒) 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑀𝑜𝑙𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 (𝑋) = 𝑎𝑚𝑜𝑢𝑛𝑡 (𝑚𝑜𝑙)𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 + 𝑎𝑚𝑜𝑢𝑛𝑡 (𝑚𝑜𝑙) 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 Example: Find the mole fractions of alcohol and water in a sample of homemade hand sanitizer that contains 58.0 grams of water and 142 grams of isopropyl alcohol (C3H7OH). Solutions: The mass and chemical formula of each component is given. This must be converted to moles for you to proceed with the calculations. 1 𝑚𝑜𝑙 𝐶3 𝐻7 𝑂𝐻 𝐴𝑚𝑜𝑢𝑛𝑡 (𝑚𝑜𝑙)𝑜𝑓 𝐶3 𝐻7𝑂𝐻 = 142 𝑔 𝐶3 𝐻7𝑂𝐻 𝑥 = 2.36 𝑚𝑜𝑙 𝐶3 𝐻7𝑂𝐻 60.09 𝑔 𝐶3 𝐻7𝑂𝐻 1 𝑚𝑜𝑙 𝐻2 𝑂 𝐴𝑚𝑜𝑢𝑛𝑡 (𝑚𝑜𝑙)𝑜𝑓 𝐻2 𝑂 = 52.0 𝑔 𝐻2 𝑂 𝑥 = 3.22 𝑚𝑜𝑙 𝐻2 𝑂 18.02 𝑔 𝐻2 𝑂 Solving for the mole fractions: 𝑎𝑚𝑜𝑢𝑛𝑡 (𝑚𝑜𝑙𝑒) 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑀𝑜𝑙𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 (𝑋) = 𝑎𝑚𝑜𝑢𝑛𝑡 (𝑚𝑜𝑙)𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 + 𝑎𝑚𝑜𝑢𝑛𝑡 (𝑚𝑜𝑙) 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐶3 𝐻7 𝑂𝐻 2.36 𝑚𝑜𝑙 𝑋 𝑜𝑓 𝐶3 𝐻7 𝑂𝐻 = = 𝑡𝑜𝑡𝑎𝑙 𝑚𝑜𝑙𝑒𝑠 2.36 𝑚𝑜𝑙 + 3.22 𝑚𝑜𝑙 𝑋 𝑜𝑓 𝐶3 𝐻7 𝑂𝐻 = 0.423 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐻2 𝑂 3.22 𝑚𝑜𝑙 𝑋 𝑜𝑓 𝐻2 𝑂 = = 𝑡𝑜𝑡𝑎𝑙 𝑚𝑜𝑙𝑒𝑠 2.36 𝑚𝑜𝑙 + 3.22 𝑚𝑜𝑙 𝑋 𝑜𝑓 𝐻2 𝑂 = 0.577 6. Proof Alcohol proof is the measure of the ethanol content of an alcoholic beverage. It is calculated by simply multiplying the volume of ethyl alcohol by 2. Examples: 40% ethyl alcohol by volume is 80 proof 70-proof whiskey is 35% alcohol by volume 88-proof whiskey is 44% alcohol by volume Chem 1 Expression of Concentration Page 8 of 12  NONGRADED ASSESSMENT Read the following items with great comprehension. Give what is asked. Show your complete solutions and box your final answer with proper units and number of significant figures. 1. Ethanol (C2H5OH) is a colorless, volatile and flammable liquid and is the intoxicating agent in your Christmas wines. A 101- L of ethanol solution was prepared by mixing 1.00 g ethanol with 100.0 g water. What will be the molarity, mass percent, mole fraction, and molality of ethanol in this solution? 2. Nitric acid (HNO3) is colorless, corrosive, a substance that gives off fumes, and is essential in the manufacture of explosives and fertilizers. You were tasked to prepare a solution containing 12.6 grams of HNO3 in 1.0-liter solution. Find its molarity. 3. You were a bit confused during your Chemistry class. The instructions clearly stated to prepare a solution involving 15.20 grams of iodine, I2, in 1.33 moles of diethyl ether, (CH3CH2)2O. You were able to accomplish the task easily Navigate because of your notes. Now, calculate its molality. GRADED ASSESSMENT 8 min The same instructions from the NONGRADED ASSESSMENT will be applied here. This will be done outside the 30-minute period. 1. You were tasked to prepare an ethanol-water solution. A volume of 10.00 mL of ethanol must be dissolved in a sufficient amount of water to produce 100.0 mL of a solution with a density of 0.982 g/ml. Find the concentration of ethanol in this particular solution expressed in different concentration units: a. volume percent b. mass percent c. mass/volume percent d. mole fraction e. mole percent f. molarity g. molality Density of ethanol = 0.789 g/ml 2. Find the molality of each of the following solutions: a. a pitcher with a mixture of 14.3 grams of table sugar (C12H22O11) and 676 grams of water Chem 1 Expression of Concentration Page 9 of 12  b. 7.20 moles of an antifreeze, ethylene glycol (C2H6O2), mixed with 3546 grams of water 3. You were given two solutions. Which of the two is more concentrated? Solution A: 500.0-mL solution with 50.0-gram CaCO3 Solution B: 4.0-L solution with 6.0-moles H2SO4 Percent by Mass: the ratio of mass of solute to mass of solution times 100% Percent by Volume: the ratio of volume of solute to volume of solution times 100% Percent by Mass/Volume: the ratio of mass of solute to volume Knot of solution times 100% Parts per Million (ppm)/Parts per Billion (ppb): used when 1 min dealing with extremely diluted solutions. Molarity (M): moles solute per liter of solution Molality (m): moles solute per mass of solvent (in kg) Mole fraction (x): ratio of moles of a given component to total moles of all components Proof: Twice the volume of ethyl alcohol in an alcoholic beverage a suggested time allocation set by the teacher b actual time spent by the student (for information purposes only) ANSWER KEYS NONGRADED ASSESSMENT 1. 𝟏 𝒎𝒐𝒍 𝑪𝟐 𝑯𝟓 𝑶𝑯 a. 𝟏. 𝟎𝟎 𝒈 𝑪𝟐 𝑯𝟓 𝑶𝑯 𝒙 = 𝟐. 𝟏𝟕 𝒙 𝟏𝟎−𝟐 𝒎𝒐𝒍 𝑪𝟐 𝑯𝟓 𝑶𝑯 𝟒𝟔.𝟎𝟕 𝒈 𝑪𝟐 𝑯𝟓 𝑶𝑯 𝟏𝑳 𝒗𝒐𝒍𝒖𝒎𝒆 = 𝟏𝟎𝟏 𝒎𝑳 𝒙 = 𝟎. 𝟏𝟎𝟏 𝑳 𝟏𝟎𝟎𝟎 𝒎𝑳 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝑪𝟐 𝑯𝟓 𝑶𝑯 𝟐. 𝟏𝟕 𝒙 𝟏𝟎−𝟐 𝒎𝒐𝒍 𝑴 𝒐𝒇 𝑪𝟐 𝑯𝟓 𝑶𝑯 = = 𝑳 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝟎. 𝟏𝟎𝟏 𝑳 𝑴 𝒐𝒇 𝑪𝟐 𝑯𝟓 𝑶𝑯 = 𝟎. 𝟐𝟏𝟓 𝑴 𝑪𝟐 𝑯𝟓 𝑶𝑯 𝒎𝒂𝒔𝒔 𝒐𝒇 𝑪𝟐 𝑯𝟓 𝑶𝑯 b. 𝒎𝒂𝒔𝒔 𝒑𝒆𝒓𝒄𝒆𝒏𝒕 𝑪𝟐 𝑯𝟓 𝑶𝑯 = 𝟏𝟎𝟎% 𝒙 𝒎𝒂𝒔𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝟏. 𝟎𝟎 𝒈 𝑪𝟐 𝑯𝟓 𝑶𝑯 𝒎𝒂𝒔𝒔 𝒑𝒆𝒓𝒄𝒆𝒏𝒕 𝑪𝟐 𝑯𝟓 𝑶𝑯 = 𝒙 𝟏𝟎𝟎% 𝟏𝟎𝟎. 𝟎 𝒈 𝑯𝟐 𝑶 + 𝟏. 𝟎𝟎 𝒈 𝑪𝟐 𝑯𝟓 𝑶𝑯 𝒎𝒂𝒔𝒔 𝒑𝒆𝒓𝒄𝒆𝒏𝒕 𝑪𝟐 𝑯𝟓 𝑶𝑯 = 𝟎. 𝟗𝟗𝟎% 𝒎𝒐𝒍𝒆𝒔 𝑪𝟐 𝑯𝟓 𝑶𝑯 c. 𝒎𝒐𝒍𝒆 𝒇𝒓𝒂𝒄𝒕𝒊𝒐𝒏 𝑪𝟐 𝑯𝟓 𝑶𝑯 = 𝒎𝒐𝒍𝒆𝒔 𝑪𝟐 𝑯𝟓 𝑶𝑯 + 𝒎𝒐𝒍𝒆𝒔 𝑯𝟐 𝑶 Chem 1 Expression of Concentration Page 10 of 12  𝟏 𝒎𝒐𝒍 𝑯𝟐 𝑶 𝒎𝒐𝒍𝒆 𝑯𝟐 𝑶 = 𝟏𝟎𝟎. 𝟎 𝒈 𝑯𝟐 𝑶 = 𝟓. 𝟓𝟔 𝒎𝒐𝒍 𝟏𝟖. 𝟎 𝒈 𝑯𝟐 𝑶 𝟐. 𝟏𝟕 𝒙 𝟏𝟎−𝟐 𝒎𝒐𝒍 𝒎𝒐𝒍𝒆 𝒇𝒓𝒂𝒄𝒕𝒊𝒐𝒏 𝑪𝟐 𝑯𝟓 𝑶𝑯 = 𝟐. 𝟏𝟕 𝒙 𝟏𝟎−𝟐 𝒎𝒐𝒍 + 𝟓. 𝟓𝟔 𝒎𝒐𝒍 𝟐. 𝟏𝟕 𝒙 𝟏𝟎−𝟐 𝒎𝒐𝒍 𝒎𝒐𝒍𝒆 𝒇𝒓𝒂𝒄𝒕𝒊𝒐𝒏 𝑪𝟐 𝑯𝟓 𝑶𝑯 = = 𝟎. 𝟎𝟎𝟑𝟖𝟗 𝟓. 𝟓𝟖 𝒎𝒐𝒍 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝑪𝟐 𝑯𝟓 𝑶𝑯 𝟐.𝟏𝟕 𝒙 𝟏𝟎−𝟐 𝒎𝒐𝒍 d. 𝑴𝒐𝒍𝒂𝒍𝒊𝒕𝒚 𝒐𝒇 𝑪𝟐 𝑯𝟓 𝑶𝑯 = = 𝟏 𝒌𝒈 𝒌𝒊𝒍𝒐𝒈𝒓𝒂𝒎 𝒐𝒇 𝑯𝟐 𝑶 𝟏𝟎𝟎.𝟎 𝒈 𝒙 𝟏𝟎𝟎𝟎 𝒈 𝟐. 𝟏𝟕 𝒙 𝟏𝟎−𝟐 𝒎𝒐𝒍 𝑴𝒐𝒍𝒂𝒍𝒊𝒕𝒚 𝒐𝒇 𝑪𝟐 𝑯𝟓 𝑶𝑯 = = 𝟎. 𝟐𝟏𝟕 𝒎 𝟎. 𝟏𝟎𝟎𝟎 𝒌𝒈 𝟏 𝒎𝒐𝒍 𝑯𝑵𝑶𝟑 2. 𝟏𝟐. 𝟔 𝒈 𝑯𝑵𝑶𝟑 𝒙 = 𝟎. 𝟐𝟎𝟎 𝒎𝒐𝒍 𝑯𝑵𝑶𝟑 𝟔𝟑.𝟎 𝒈 𝑯𝑵𝑶𝟑 𝒎𝒐𝒍 𝑯𝑵𝑶𝟑 𝟎. 𝟐𝟎𝟎 𝒎𝒐𝒍 𝑯𝑵𝑶𝟑 𝑴𝒐𝒍𝒂𝒓𝒊𝒕𝒚 = = = 𝟎. 𝟐𝟎𝟎 𝑴 𝑳 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝟏. 𝟎 𝑳 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝟏 𝒎𝒐𝒍 𝑰𝟐 3. 𝑨𝒎𝒐𝒖𝒏𝒕 (𝒎𝒐𝒍) 𝒐𝒇 𝑰𝟐 = 𝟏𝟓. 𝟐𝟎 𝒈 𝑰𝟐 𝒙 = 𝟓. 𝟗𝟖𝟗 𝒙 𝟏𝟎−𝟐 𝒎𝒐𝒍 𝑰𝟐 𝟐𝟓𝟑.𝟖 𝒈 𝑰𝟐 𝑴𝒂𝒔𝒔 (𝒌𝒈) 𝒐𝒇 (𝑪𝑯𝟑 𝑪𝑯𝟐 )𝟐 𝑶 𝟕𝟒. 𝟏𝟐 𝒈 (𝑪𝑯𝟑 𝑪𝑯𝟐 )𝟐 𝑶 𝟏 𝒌𝒈 = 𝟏. 𝟑𝟑 𝒎𝒐𝒍 (𝑪𝑯𝟑 𝑪𝑯𝟐 )𝟐 𝑶 𝒙 𝒙 𝟏 𝒎𝒐𝒍 (𝑪𝑯𝟑 𝑪𝑯𝟐 )𝟐 𝑶 𝟏𝟎𝟎𝟎 𝒈 = 𝟗. 𝟖𝟔 𝒙 𝟏𝟎−𝟐 𝒌𝒈 (𝑪𝑯𝟑 𝑪𝑯𝟐 )𝟐 𝑶 𝟓. 𝟗𝟖𝟗 𝒙 𝟏𝟎−𝟐 𝒎𝒐𝒍 𝑴𝒐𝒍𝒂𝒍𝒊𝒕𝒚 𝒐𝒇 𝑪𝟐 𝑯𝟓 𝑶𝑯 = = 𝟎. 𝟔𝟎𝟕 𝒎 𝟗. 𝟖𝟔 𝒙 𝟏𝟎−𝟐 𝒌𝒈 Chem 1 Expression of Concentration Page 11 of 12 References: Albarico, J. (PSHS-CBRZ). (n.d.). THINK Framework. Based on Science Links by E.G. Ramos and N. Apolinario, Quezon City. Rex Bookstore. Bewick, S., Parsons, R., Forsythe, T., Robinson, S. and Dupon J. (2020) CK-12 Foundation. Retrieved December 24, 2020 from Chem LibreTexts. Chang, R. (2010). Chemistry. 10th Edition. New York, USA: McGraw-Hill Education. Petrucci, R. (2011). General Chemistry: Principles and Modern Application. 10th Edition. Pearson Canada Inc., Toronto, Ontario. Silberberg, M. (2015). Chemistry: The Molecular Nature of Matter and Change. 7th Edition. New York, USA: McGraw-Hill Education. Zumdhal, S. (2018). Chemistry. 10th Edition. Boston, USA: Cengage Learning. Prepared by: Jezreel A. Gaa Reviewed by: Lester Mendoza Position: Special Science Teacher (SST) – I Position: Special Science Teacher (SST) - II Campus: MIMAROPA Region Campus: Main © 2020 Philippine Science High School System. All rights reserved. This document may contain proprietary information and may only be released to third parties with approval of management. Document is uncontrolled unless otherwise marked; uncontrolled documents are not subject to update notification. Chem 1 Expression of Concentration Page 12 of 12