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# CHEMISTRY OF SOLUTIONS LECTURE 1 NOTES 202-NYB-05

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This 4 page Class Notes was uploaded by CatLover44 on Friday September 16, 2016. The Class Notes belongs to 202-NYB-05 at Dawson Community College taught by Nadia Schoonhoven in Fall 2016. Since its upload, it has received 9 views. For similar materials see Chemistry of Solutions in Chemistry at Dawson Community College.

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Date Created: 09/16/16

Chemistry of Solutions Course Number: 202-NYB-05 Lecture no. 1 Date: Thursday, August 25, 2016 Professor: Nadia Schoonhoven Topics Covered: Solution composition: a short review of significant figures (students are graded for their use of significant figures, further abbreviated as “sig figs”); review of mass percent, the mole fraction, molarity, molality and parts per million (ppm). How to Count Sig Figs Do your best to memorize these as marks will be deducted if you use the incorrect number of sig figs in your calculations! Scenario no. 1: when a number doesn’t contain a decimal or any zeros, all of the digits count as sig figs. For example, 1234 has 4 sig figs. Scenario no. 2: when zeroes precede nonzero numbers, only the nonzero digits count as sig figs. For example, 0.01 has 1 sig fig. Scenario no. 3: zeroes that are sandwiched between nonzero digits always count as sig figs. For example, 20.04 has 4 sig figs. Scenario no. 4: zeroes that follow nonzero digits are only significant if there is a decimal in that number. For example, 1.200 has 4 sig figs. Scenario no. 5: exact numbers (e.g.: data obtained from a laboratory experiment) are counted, not measured. These numbers have an infinite number of sig figs. For example, there are exactly 2.54 cm in one inch. Using Sig Figs in Calculations There are certain rules to follow when counting sig figs while doing calculations. • If you’re multiplying or dividing, your final answer should have the same amount of digits as the number with the least number of sig figs. For example, 1.202 (4 sig figs) ÷ 2.1 (2 sig figs) = 0.572381 (7 sig figs), final answer: 0.6 (2 sig figs). • If you're adding or subtracting, your final answer should have the same amount of decimal spaces as the value with the least amount of decimal spaces. For example, 0.12 (2 decimal spaces) + 1.123 (3 decimal spaces) = 1.243 (3 decimal spaces), final answer: 1.24 (2 decimal spaces) Solution Composition • Solutions are homogeneous mixtures made up of at least two components. These components can be liquids, gases, or solids. • While homogeneous solutions are solutions that are only made of one chemical compound, a homogeneous mixture is a solution with uniform consistency. One example of a homogeneous mixture is salt stirred into a glass of water. • When we discuss solutions, we must consider solutes and solvents. This course mainly deals with liquid solutions. o A solution is comprised of solutes and solvents; o The solute in a solution is the substance that is being dissolved; o The solvent in a solution is what is doing the dissolving. Water is the universal solvent. • Ask yourself two important questions when solving problems involving homogenous solutions: 1) How much solute is being dissolved? 2) What conditions affect the solubility of the solution? o Recall: the solubility of a given solution refers to a solute’s ability to dissolve in a solvent. o Recall: like substances dissolve like substances (“like dissolves like”). This is so because of the strength of their intermolecular forces (i.e. London dispersion forces, dipole-dipole forces, etc.). Increasing the strength of intermolecular forces increases the amount of interaction between molecules. Units of Concentration 1) Fractional Composition (sum of all the individual parts of a solution equals the whole solution). a) Calculating Mass Percent: To calculate the mass percent of a solution, divide the given mass of the solute by the mass of the whole solution, then multiply this value by 100. Mass % = ( Mass of Solute ÷ Mass of Solution) × 100 Note: the sum of all the individual solutes divided by the mass of the solution should equal to 1. For instance, (Mass of Solute A/Mass of Solution) + (Mass of Solute B/Mass of Solution) + (Mass of Solute C/Mass of Solution) = 1. The mass percent gives us the mass of a solute per total mass of solution. b) Calculating Mole Fraction: To calculate the mole fraction, find the number of moles of the solute, and divide this number by the total number of moles in the solution. Calculate the total number of moles of solution by adding up all the moles of the individual solutes in the solution. Mole Fraction = ( Moles of Solute A ) ÷ ( Moles of Solute A + Moles of Solute B + Moles of Solute C) Note: the sum of all the moles of solutes in the solution should equal to 1, which means Moles of Solute A + Moles of Solute B + Moles of Solute C = 1. 2) Other Units of Concentration c) Molarity: Molarity is defined by moles of solute per litre of solution. This unit of concentration is temperature-dependent because it is based on the volume of a solution. This value represents the volume of solute + the volume of solvent. Molarity = M = Mole/Litre d) Molality: Molality is defined by moles of solute per kilogram of solvent. This unit of concentration is not temperature-dependent because mass is unaffected by changes in temperature. Molality is denoted by a lowercase cursive “m”. Molality = m = Moles of solute/Kilograms of solution. e) Parts per million: Parts per million (ppm) is defined by the mass of a chemical per litre. Ppm and mg/L are interchangeable units of concentration; they mean the same thing. Exercises From Lecture Slides & Their Solutions 1) A solution of sodium chloride was made by dissolving 7.00 g of NaCl in 50.0 mL of water. Calculate the mole fraction of sodium chloride, assuming that water has a density of 1.00 g/mL. Step 1: calculate the molar mass of NaCl: Molar Mass of NaCl = (22.99 g) + (35.45 g) = 58.44 g All the values have the same number of decimal spaces, so we don't need to round up or down. Step 2: calculate the number of moles of NaCl in 7.00 g: 1 mol. NaCl = 58.44 g X mol. NaCl = 7.00 g To find the value of X: what you want goes on top. 7.00 g NaCl x 1 mol. NaCl ÷ 58.44 g NaCl = 0.1198 mol. NaCl. Step 3: calculate the molar mass of water. Molar Mass of H2O = (2 x 1.008 g) + (15.99 g) = 18.006 g. Step 4: calculate the number of moles of water in 50.0 mL using its density. (50.0 mL H O2 x (1.00 g H O)/2mL) x (mol. H O)/(12.006 g) = 2.777 mol. H O. 2 Step 5: calculate the mole fraction: m = mol. of solute/ mol. of solution m = (0.1198 mol. NaCl)/ (2.777 mol. H O + 0.1198 mol. NaCl) = 0.04136 mol. 2 Final answer with correct number of sig figs: 0.0414 mol. (4 sig figs). 2) A solution of sodium chloride was made by dissolving 7.00 g (3 sig figs) of NaCl with 50.0 mL (3 sig figs) of water. Calculate the molality of sodium chloride, assuming that water has a density of 1.00 (3 sig figs) g/mL. Step 1: convert 50.0 mL of water to 50.0 g of water, then convert grams to kilograms. What you want goes on top. 50.0 mL H2O x (1.00 g/ mL) = 50.0 g of H2O 50.0 g/ 1000 mL = 0.05 kg. Step 2: Use the number of moles of NaCl in 7.00 g (0.11978 mol. NaCl in 7.00 g) and the mass of water in kilograms to calculate m. m = mol. of solute/ kg of solution m = (0.11978 mol.)/ (0.05 kg) = 2.3956 mol./ kg. Final answer with the right number of sig figs: 2.40 mol./kg (3 sig figs).

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