 # CONCENTRATION OF SOLUTIONS Day 4 CONCENTRATION  A measurement of the amount of solute in a given amount of solvent or solution (unit of measurement.

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CONCENTRATION OF SOLUTIONS Day 4

CONCENTRATION  A measurement of the amount of solute in a given amount of solvent or solution (unit of measurement = molarity )  Kind of like the “strength” of the solution

MOLARITY Molarity (M)= mol solute L solution

VOLUME CONVERSION FACTORS  1 cm 3 = 1 mL  1 dm 3 = 1 L

DILUTE  A dilute solution is a solution in which there is more solvent than solute.  For example, to dilute some solution, you could add excess water to it.

EXAMPLE #1: If we put 0.500 mol of sodium hydroxide into 1.00 L of solution, what is the molarity of the solution?

Molarity= mol solute/L sol’n M= 2.00 L 0.50 mol NaOH = 0.25 M NaOH

Example #2: What is the molarity if we have 80.0 g NaOH in 1.00 L of solution?

Molarity= mol solute/L sol’n M= 1.00 L 2.00 mol NaOH = 2.00 M NaOH 80.0 g NaOH 40.0 g NaOH 1 mol NaOH =2.00 mol

EXAMPLE #3:  How many moles of KOH are in 0.50 L of 0.10 M KOH?

Molarity= mol solute cross multiply! 1 L sol’n mol KOH= (0.50 L) (0.10 M KOH) = 0.05 mol KOH

EXAMPLE #4  How many grams of KOH are in 2.75 L of 0.25 M KOH?

Molarity = moles cross multiply! L Moles = Molarity * L = 0.25 mol/L * 2.75L = 0.69 moles KOH 0.69 mol KOH 56.097 g KOH = 38.7 g KOH 1 mole KOH (39)

MOLARITY - DILUTION Some chemicals are sold as pre-prepared concentrated solutions (stock solutions). To be used, stock solutions usually must be diluted.

A LWAYS A DD A CID

MOLARITY – DILUTION CALCULATIONS M 1 V 1 = M 2 V 2 M 1 – Molarity of stock solution V 1 – Volume of stock solution (L or mL) M D – Molarity of dilute solution V D – Volume of dilute solution (L or mL) Stock (original)Dilute (new solution)

*In dilution calculations, the units for volume must be the same.

2.) MOLARITY - DILUTION  Example #1: How much 12.0 M HCl is required to make 2.50 L of a 0.500 M solution?  M s * V s = M D * V D

M S V S = M D V D OR M 1 V 1 = M 2 V 2 M S = 12.0 M HCl M D = 0.500 M HCl V S = ?? V D = 2.50 L V S = (0.500 M HCl) (2.50 L) (12.0 M HCl) = 0.104 L HCl

2.) MOLARITY - DILUTION  Example #2: What volume of a 1.50 M solution can be made using 0.0250 L of 18.0 M H 2 SO 4 ?  M S V S = M D V D

M S V S = M D V D M S = 18.0 M H 2 SO 4 M D = 1.50 M H 2 SO 4 V S =.0250 L V D = ???? V D = (18.0 M ) (18.0 M H 2 SO 4 ) (.0250 L) (1.50 M ) (1.50 M H 2 SO 4 ) = 0.300 L 0.300 L H 2 SO 4

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