Concentration+of+Solutions

V   i  n n y Definition: the amount of solute dissolved in a solvent

REMEMBER: if you add a small amount of ethanol to a large amount of water, the solvent is water. //However//, adding a small amount of water to a large amount of ethanol makes ethanol the solvent!

** Percent Composition by Mass (%) ** is the mass of the solute divided by the mass of the solution, multiplied by 100 to get the percentage. To calculate the mass of the solution, you add together the mass of the solute and the mass of the solvent.


 * Mole Fraction (X) ** is when you take the number of moles in a compound divided by the total moles of all compounds in a solution (Hint: the total mole fractions in a solution equals 1!).


 * Molarity** is the moles of a solute divided by the liters of the solution.


 * Molality** is the moles of a solute divided by the kilograms of the solvent.

Molarity depends on the volume of the solution, while molality depends on the mass of the solvent. Also, because molality deals with mass, it isn't affected by temperature. Molarity has to vary with temperature because the expansion/contraction of a solution changes its volume, so molality is used when working with a range of temperatures. saddlebackphysio.blogspot.com
 * Difference between Molarity and Molality!**

Don't use this mole! > > 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 solutio **Volume Percent (% v/v)** Volume percent or volume/volume percent most often is used when preparing solutions of liquids. Volume percent is defined as: > Note that volume percent is relative to volume of solution, not volume of //solvent//. For example, wine is about 12% v/v ethanol. This means there are 12 ml ethanol for every 100 ml of wine. It is important to realize liqud and gas volumes are not necessarily additive. If you mix 12 ml of ethanol and 100 ml of wine, you will get less than 112 ml of solution. > As 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). > Example: > > 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) > 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 = 5 + 1 = 6 mol > > x water =5 mol / 6 mol= 0.833 > > x glycerol =1 mol / 6 mol= 0.167 > > It's a good idea to check your math by making sure the mole fractions add up to 1: > > x water + x glycerol =.833 + 0.167= 1.000 > Example: > > What is the molarity of a solution made when water is added to 11 g CaCl 2 to make 100 mL of solution? > Solution: > > 11 g CaCl 2 / (110 g CaCl 2 / mol CaCl 2 ) = 0.10 mol CaCl 2 > > 100 mL x 1 L / 1000 mL = 0.10 L > > molarity = 0.10 mol / 0.10 L > > molarity = 1.0 M > Example: > > What is the molality of a solution of 10 g NaOH in 500 g water? > Solution: > > 10 g NaOH / (4 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 > Example: > > 1 M sulfuric acid (H 2 SO 4 ) 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.
 * 1) Example:
 * 1) v/v % = [(volume of solute)/(volume of solution)] x 100%
 * 1) **Mole Fraction (X)** This is the number of moles of a compound divided by the total number of moles of all chemical species in the solution. Keep in mind, the sum of all mole fractions in a solution always equals 1.
 * 1) **Molarity (M)** Molarity is probably the most commonly used unit of concentration. It is the number of moles of solute per liter of solution (not necessarily the same as the volume of solvent!).
 * 1) **Molality (m)** 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.
 * 1) **Normality (N)** 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 capcity of a given molecule. Normality is the only concentration unit that is reaction dependent.

http://chemistry.about.com/od/lecturenotesl3/a/concentration.htm

>> Calculating Dilution

>> Dilution calculations are simplified by using the following equation:

>> M1V1 = M2V2

Where: M1 = concentration of the first solution V1 = volume of the first solution M2 = concentration of the second solution V2 = volume of the second solution

>> Concentration and volume in the equation above can have any units as long as the units are the same for the two solutions.

As long as you know three of the four values from the equation above, you can calculate the fourth. Let's consider a sample problem: >>> You have 1 L of a 0.125 M aqueous solution of table sugar. You want to dilute the solution to 0.05 M. What do you do?

>>

>> To solve the problem, you simply plug in the three numbers you know:

>> (0.125 M) (1 L) = (0.05 M) V2 >> 2.5 L = V2

>> Using the equation, you determine that the volume of the diluted solution should be 2.5 L. So we simply add enough water to the first solution so that the solution's volume becomes 2.5 L.