II. Solution Concentration (p. 480 – 486)

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Presentation transcript:

II. Solution Concentration (p. 480 – 486) Ch. 16 – Solutions II. Solution Concentration (p. 480 – 486)

A. Concentration The amount of solute in a solution Describing Concentration % by mass - medicated creams % by volume - rubbing alcohol ppm, ppb - water contaminants molarity - used by chemists molality - used by chemists

substance being dissolved B. Percent Solutions Percent By Volume (%(v/v)) Concentration of a solution when both solute and solvent are liquids often expressed as percent by volume substance being dissolved total combined volume

B. Percent Solutions Find the percent by volume of ethanol (C2H6O) in a 250 mL solution containing 85 mL ethanol. Solute = 85 mL ethanol Solution = 250 mL 85 mL ethanol 250 mL solution % (v/v) = = 34% ethanol (v/v) x 100

substance being dissolved B. Percent Solutions Percent By Mass (%(m/m)) Concentration of a solution when solute is solid sometimes expressed as percent by mass substance being dissolved total combined volume

B. Percent Solutions How much glucose should you use to make 2000. g of a 2.8% (m/m) solution in water? Percent (m/m) = 2.8% Solution = 2000. g unknown g glucose 2000. g solution 2.8 = = 56.00 g glucose x 100

substance being dissolved C. Molarity Concentration of a solution most often used by chemists substance being dissolved total combined volume

C. Molarity 2M HCl What does this mean?

D. Molarity Calculations LITERS OF GAS AT STP Molar Volume (22.4 L/mol) MASS IN GRAMS MOLES NUMBER OF PARTICLES Molar Mass (g/mol) 6.02  1023 particles/mol Molarity (mol/L) LITERS OF SOLUTION

D. Molarity Calculations How many moles of NaCl are required to make 0.500L of 0.25M NaCl? 0.500 L sol’n 0.25 mol NaCl 1 L sol’n = 0.013 mol NaCl

D. Molarity Calculations How many grams of NaCl are required to make 0.500L of 0.25M NaCl? 0.500 L sol’n 58.44 g NaCl 1 mol NaCl 0.25 mol NaCl 1 L sol’n = 7.3 g NaCl

D. Molarity Calculations Find the molarity of a 250 mL solution containing 10.0 g of NaF. 10.0 g NaF 1 mol NaF 41.99 g NaF = 0.95 M NaF .25 L sol’n

E. Dilution Preparation of a desired solution by adding water to a concentrate Moles of solute remain the same

E. Dilution What volume of 15.8M HNO3 is required to make 250 mL of a 6.0M solution? GIVEN: M1 = 15.8M V1 = ? M2 = 6.0M V2 = 250 mL WORK: M1 V1 = M2 V2 (15.8M) V1 = (6.0M)(250mL) V1 = 95 mL of 15.8M HNO3

F. Molality mass of solvent only 1 kg water = 1 L water

G. Molality Calculations Find the molality of a solution containing 75 g of MgCl2 in 250 mL of water. 75 g MgCl2 1 mol MgCl2 95.21 g MgCl2 0.25 kg water = 3.2m MgCl2

G. Molality Calculations How many grams of NaCl are req’d to make a 1.54m solution using 0.500 kg of water? 0.500 kg water 1.54 mol NaCl 1 kg water 58.44 g NaCl 1 mol NaCl = 45.0 g NaCl

H. Preparing Solutions 1.54m NaCl in 0.500 kg of water 500 mL of 1.54M NaCl mass 45.0 g of NaCl add water until total volume is 500 mL mass 45.0 g of NaCl add 0.500 kg of water 500 mL water 45.0 g NaCl 500 mL mark volumetric flask

H. Preparing Solutions Copyright © 1995-1996 NT Curriculum Project, UW-Madison (above: “Filling the volumetric flask”)

H. Preparing Solutions Copyright © 1995-1996 NT Curriculum Project, UW-Madison (above: “Using your hand as a stopper”)

H. Preparing Solutions 250 mL of 6.0M HNO3 by dilution measure 95 mL of 15.8M HNO3 95 mL of 15.8M HNO3 combine with water until total volume is 250 mL 250 mL mark Safety: “Do as you oughtta, add the acid to the watta!” or AA – add acid! water for safety

Solution Preparation Mini-Lab Turn in one paper per team. Complete the following steps: A) Show the necessary calculations. B) Write out directions for preparing the solution. C) Prepare the solution. For each of the following solutions: 1) 100.0 mL of 0.50M NaCl 2) 0.25m NaCl in 100.0 mL of water 3) 100.0 mL of 3.0M HCl from 12.0M concentrate. (don’t actually prepare this one!)