13.3 Concentration,Molarity and Molality

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13.3 Concentration of Solutions:

13.3 Concentration of Solutions

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The concentration of a solution is a measure of the amount of solute in a given amount of solvent or solution

Molarity:

Molarity Molarity (M) is the number of moles of solute in one liter of solution. Molarity(M) = amount of solute (mol) volume of solution (L ) **pay close attention to the units involved in this calculation. If the given is not in moles and liters, you will have to convert before you plug into the equation.

Molarity Example #1:

Molarity Example #1 You have 3.50 L of a solution that contains 90.0 g of sodium chloride, NaCl . What is the molarity of that solution? 90.0 g NaCl 1 mol NaCl = 1.54 mol NaCl 58.44 g NaCl M = mol solute 1.54 mol NaCl = .440 M NaCl L solution 3.50 L

Molarity example #2:

Molarity example #2 You have .8 L of a .5 M HCl solution. How many moles of HCl does this solution contain? M = mol solute .5 M HCl = X mol L solution .8 L = .4 mol HCl

Molality:

Molality Molality (m) is the concentration of a solution expressed in moles of solute per kilogram of solvent. m = moles solute mass of solvent (kg ) ** remember to convert to moles and kg if necessary

Molality example :

Molality example A solution was prepared by dissolving 29.8 g of sucrose (C 12 H 22 O 11 ) in 125 g of water. Find the molal concentration of this solution. m = moles solute mass of solvent (kg) 29.8 g C 12 H 22 O 11 mol C 12 H 22 O 11 = .0870 mol C 12 H 22 O 11 342.34 g C 12 H 22 O 11 m = .0870 mol = .696 m C 12 H 22 O 11 .125 kg

Diluting Solutions:

Diluting Solutions Not all compounds are in the solid form. Acids are purchased as liquids (“stock solutions”). Yet, we still need a way to make molar solutions of these compounds. We use the formula: M 1 V 1 = M 2 V 2 M 1 and V1 represent the initial Molarity and volume and M 2 and V 2 represent the desired molarity and volume

Dilution example:

Dilution example If we have 1 L of 3 M HCl , what is the molarity if we dilute acid to 6 L? M 1 = 3M V 1 = 1L M 2 = X V 2 = 6L M 1 V 1 = M 2 V 2 (3M) (1L) = (X) (6 L) X = .5 M

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