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**Percent By Mass Percent by mass, %m/m can be calculated as:**

Defined as how many parts exist in every 100 parts of the whole; In other words, the mass of the solute divided by the mass of the solution (mass of solute plus mass of solvent), multiplied by 100. Percent by mass, %m/m can be calculated as: Example: Determine the percent composition by mass of a 100 g salt solution which contains 20 g salt. Solution: 20 g NaCl / 100 g solution x 100 = 20% NaCl solution

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Percent By Volume If both the solute and solvent are both liquids, we need to use a different way to define the concentration percentage. Volume percent or volume/volume percent most often is used when preparing solutions of liquids. Another example: 70% v/v rubbing alcohol may be prepared by taking 700 ml of isopropyl alcohol and adding sufficient water to obtain 1000 ml of solution (which will not be 300 ml).

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Mole Fraction Yet another way to use the famous Mole, Mole Fraction takes into account the number of moles of a particular compound compared to the number of total moles in solution. Ex.: What are the mole fractions of the components of the solution formed when 92 g glycerol is mixed with 90 g water? (molecular weight water = 18; molecular weight of glycerol = 92)

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**Mole Fraction Example:**

Solution: 90 g water = 90 g x 1 mol / 18 g = 5 mol water 92 g glycerol = 92 g x 1 mol / 92 g = 1 mol glycerol total mol = = 6 mol xwater = 5 mol / 6 mol = x glycerol = 1 mol / 6 mol = It's a good idea to check your math by making sure the mole fractions add up to 1: xwater + xglycerol = = 1.000

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Molality Molality is the number of moles of solute per kilogram of solvent. Because the density of water at 25°C is about 1 kilogram per liter, molality is approximately equal to molarity for dilute aqueous solutions at this temperature. This is a useful approximation, but remember that it is only an approximation and doesn't apply when the solution is at a different temperature, isn't dilute, or uses a solvent other than water. Ex.: What is the molality of a solution of 10 g NaOH in 500 g water?

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Solution 10 g NaOH / (40 g NaOH / 1 mol NaOH) = 0.25 mol NaOH 500 g water x 1 kg / 1000 g = 0.50 kg water molality = 0.25 mol / 0.50 kg molality = 0.05 M / kg molality = 0.50 m

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Normality Normality is equal to the gram equivalent weight of a solute per liter of solution. A gram equivalent weight or equivalent is a measure of the reactive capacity of a given molecule. Normality is the only concentration unit that is reaction dependent. Ex.: 1 M sulfuric acid (H2SO4) is 2 N for acid-base reactions because each mole of sulfuric acid provides 2 moles of H+ ions. On the other hand, 1 M sulfuric acid is 1 N for sulfate precipitation, since 1 mole of sulfuric acid provides 1 mole of sulfate ions.

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SOLUTIONS Chapter 15.

SOLUTIONS Chapter 15.

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