# II. Solution Concentration (p. 480 – 486)

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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 g of a 2.8% (m/m) solution in water? Percent (m/m) = 2.8% Solution = g unknown g glucose 2000. g solution 2.8 = = 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 = 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 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 kg of water 500 mL water 45.0 g NaCl 500 mL mark volumetric flask
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