oz ounce (mass) = L = cm 3

Transcription

1 Memorize relationships shown in each box! NOTE: Exact quantities are specified as exact. Also, consider 1 as exact. mass (M) Common unit abbreviations (singular) 1 kg = lb m = oz L liter (volume) 1 lb m = 16 oz (exact)= g gal gallon (volume) length (L) 1 m = 100 cm = qt quart (volume) (exact) angstroms (Å) fl-oz fluid-ounce (volume) = in = ft = yd = mi yd yard (length) 1 ft = 12 (exact) in = 1/3 (exact) yd = m volume (L 3 m meter (length) ) 1m 3 = 1000 L = 10 6 mi mile (length) ml = ft 3 in inch (length) = gal = qt 1 L = 1000 cm 3 ft feet (length) = 1000 ml 1 ft 3 = 1728 (exact) in 3 = gal = m 3 oz ounce (mass) = L = cm 3 lb pound (mass) (mass) g gram (mass) force (mass x acceleration) 1 N = 1 kg.m/s 2 = 10 5 dyn = 10 5 g.cm/s 2 hr, min, s hour, min, s (time) = lb f 1 lb f = lb m.ft/s 2 = N = x10 5 atm atmosphere (pressure) dynes Pa Pascal (pressure) pressure (Force/area) Absolute (P a)= Gauge (P g) + atmospheric (P torr torr (pressure) atm) 1 atm = 760 (exact) mmhg (note: torr = mmhg) mmhg millimeters mercury (pressure) = x10 5 N/m 2 (Pa) = kpa = bar psia lbf/in 2 (absolute) (pressure) = m H 2O at 4 C = lb f/in 2 (psi) psig lbf/in 2 (gauge) (pressure) = ft H 2O at 4 C J joule (energy) = in Hg at 0 C cal calorie (energy) energy (Force x distance) kwh kilowatt-hour (energy) 1 J = 1 N.m = cal Btu British thermal units (energy) = 2.778x10-7 kw.hr = ft.lb f = 9.486x10-4 Btu ev electron volt (energy) 1 cal = J = 10-3 Cal (food calories) Cal food calorie (energy) 1 ev = x10-19 J N Newton (force) 1 V = 1 J/C lb f pound (force) (force) 1 L.atm = J hp horse power (power) power (Energy/time) W watt (power) 1 W = 1 J/s = cal/s = ft.lb f/s mol mole (count) = 9.486x10-4 Btu/s = 1.341x10-3 hp ct count (count) Memorize these metric system prefixes K Kelvin (temperature) (Abbreviations are case sensitive!) C deg. celsius (temperature) Prefix abbreviation value F deg. farenheit (temperature) pico p nano n 10-9 Useful constants micro μ 10-6 g = m/s 2 gravitational acceleration milli m 10-3 c = x10 8 m/s speed of light in vacuum centi c 10-2 R= L.atm/(mol.K) gas constant deci d 10-1 R= J/(mol.K) gas constant deka da 10 1 hecto h 10 2 kilo k 10 3 mega M 10 6 giga G 10 9 tera T e = x10-19 C m e = x kg h = x10-34 J.s/photon F = C/mol e - N = 6.022x10 23 ct/mol Page 1 of , K. Golestaneh (revision 2018) electron charge electron rest mass Planck s constant Faraday s Constant Avogadro s number

2 You can detach the previous page and keep in your binder for quick reference while doing your homework problems. Remember to memorize and recall boxed quantities. Page 2 of , K. Golestaneh (revision 2018)

3 Review these pages and do the example problems listed. Show your work on the spaces provided in an organized manner with all logical steps and units shown. Box the final answer(s) and report to correct units and significant figures. These problems are collected for grading. Solutions and Concentration Units Chemists, Biologists, Physicians, Nurses and Nutritionists often deal with concentration of a pure compound in a homogenous mixture. We live in a world of matter that is complex and multicomponent. All matter around us is impure and exists in homogeneous or heterogeneous physical states. Concentration refers to the component(s) of interest in in impure matter. How do we report concentration of a component in a solution? What makes a solution? When a solute and solvent mix homogenously (ex. two miscible liquids), it is often due to the fact that they have similar intermolecular forces of attraction between their particles (atoms, molecules, ions) Concentration is commonly defined for a particular solute: Concentration (solute A) = Solute A Quantity Solution (mixture) Quantity Molarity (M) is very commonly used in chemistry especially when we deal with chemical reactions. (Let s use sln as the abbreviation for solution), Molarity = or M =,, Molality (m) is also used in chemistry especially when we deal with and account for colligative properties of solutions. These are properties which change with the moles of solute such as freezing point and boiling point of a solution as compared to freezing point and boiling point of the solvent., Molality = or m =,, Percent is a common concentration unit. Percent literally means per 100. We can refer to percent at parts of solute per 100 parts of solution, but we have to define the solute and solution parts based on volume (v/v), mass (m/m) or mole (mol/mol) units. Percent is dimensionless as the numerator and denominator units cancel. The following are other commonly used concentration units. mass % (solute) = volume % (solute) = mol % (solute) = ( ) ( ) ( ) ( ) ( ) ( ) x 100 x 100 x 100 Don t forget to use matching units of mass or matching units of volume or mass when using the above relationships. Percent has no units! The 100 multiplier is exact! Here are other concentration terms that are similarly defined (like percent) except that they are expressed as per million or per billion. Page 3 of , K. Golestaneh (revision 2018)

4 Parts per million (ppm) and parts per billion (ppb) are used when expressing very low (yet important!) level of solute concentration using volume or mass units (units of mass or volume must be matching!). Volume units are used when solute is a pure liquid (ex. alcohol, acetone, hydrocarbons, etc.) ppm (by mass) is defined as mass units of the solute in 10 6 mass units of the solution. ppb (by mass) is defined as mass units of the solute in 10 9 mass units of the solution. 1 ppm (solute) = ppm (solute) = mass (solute) mass (sln) x ppb (solute) = ppb (solute) = mass (solute) mass (sln) x 10 9 Problem 1: A solution contains mg of mercury (Hg) in a total solution mass of 75 kg, a) Calculate concentration of Hg in both ppm and ppb units. (Don t forget to keep the mass units the same!). Show your work! Hint: Essentially you want to calculate grams of Hg in 10 6 g of solution (given). This calculation work parallels percent which is calculating mass of a solute in 100 g of solution (given)!! Think about that! b) Calculate concentration of Hg in ppb units. c) Which is a better unit to use? (ppb or ppm) why? Please note that when we deal with a solution composed of very dilute solute added to water, we can safely ignore the mass of solute in the denominator and just use the density of water value of 1.00 g/ml and express a new and convenient (from lab point of view) unit for ppm by converting the mass of solution to the volume (liter) of solution. We measure volumes in lab very commonly in ml and L units. Therefore, the new unit that is equivalent to ppm (by mass) is (memorize these relationships): ppm (solute) = g solute mg solute 10 = = g sln L sln μg solute ml sln Page 4 of , K. Golestaneh (revision 2018)

5 Problem 2: Verify the above relationship by converting 25.7 ppm Na (by mass) to mg/l concentration unit. Show your work! Using the same line of reasoning we can concluded that: g solute μg solute ppb (solute) = 10 = = g sln L sln ng solute ml sln When working with dimensional analysis (unit analysis), you must translate compound units (%, M, m, ppm, ppb) to their individual units and descriptions correctly as numerator and denominator are different! For example, if a problem states 12.5 % NaCl solution (by mass): It should be translated as: 12.5 g NaCl 100 g sln If a problem states 7.2 ppm (by mass) lead (Pb) in a solution it translates as (remember the point proven in the previous example!): (Use whichever ratio that is more useful to solve the problem!): 7.2 g Pb 7.2 mg Pb 10 = g sln L sln Problems 3-6: 3) A solution should be prepared as 3.56 M calcium chloride, CaCl2. Calculate mass (g) of the solute that is needed in order to prepare 3500 ml of the solution. 4) A solution is prepared by dissolving 65 g of ethyl alcohol (density=0.78 g/ml) in 2.35 L of water. (ethyl alcohol formula: CH3CH2OH). Assume that volumes of both solute and solvent are additive which means there is no net change in volume when the two volumes are mixed. The total volume is simply the sum of individual volumes mixed together. a. Calculate the solute concentration in percent (by volume) units. Page 5 of , K. Golestaneh (revision 2018)

6 b. Calculate the solute concentration in percent (by mass) units. c. Calculate the solute concentration in molarity units. d. Calculate the mol% of solute assuming that the solute is ethyl alcohol (CH3CH2OH). 5) If seawater contains gold (Au) at 1.75 ppb by mass. What volume (gal) of seawater must be processed in a factory to obtain $55,700 value in pure gold assuming pure gold is worth $1320 per oz. Assume that seawater has a density of 1.10 g/ml. Page 6 of , K. Golestaneh (revision 2018)

7 6) Following pictures show types of volumetric glassware and pipets (and sizes) available in a chemistry lab. Imagine you work in chemistry lab and your supervisor needs you to prepare 470. ml of a sulfuric acid solution close to 0.18 M but know to 4 significant figures. You have 3500 ml of a standardized stock solution (more concentrated) of M sulfuric acid and plenty of deionized water available. Which glassware from each picture would you select to accomplish this task? Report the molarity of the prepared solution. Page 7 of , K. Golestaneh (revision 2018)

8 Check your answers: 1) a).0021 ppm b) 2.1 ppb c) ppb 2) 25.7 mg Na/L sln 3) 1400 g 4) a) 3.4% b) 2.7% c).58 M d) 1.1 % 5) 1.64E8 gal SW 6).1830 M Page 8 of , K. Golestaneh (revision 2018)