Concentration Units for AP Chemistry (page 2)
Practice problems for these concepts can be found at:
- Solutions and Colligative Properties Multipel Choice Review Questions for AP Chemistry
- Solutions and Colligative Properties Free-Response Questions for AP Chemistry
There are many ways of expressing the relative amounts of solute(s) and solvent in a solution. The terms saturated, unsaturated, and supersaturated give a qualitative measure, as do the terms dilute and concentrated. The term dilute refers to a solution that has a relatively small amount of solute in comparison to the amount of solvent. Concentrated, on the other hand, refers to a solution that has a relatively large amount of solute in comparison to the solvent. However, these terms are very subjective. If you dissolve 0.1 g of sucrose per liter of water, that solution would probably be considered dilute; 100 g of sucrose per liter would probably be considered concentrated. But what about 25 g per liter—dilute or concentrated? In order to communicate effectively, chemists use quantitative ways of expressing the concentration of solutions. Several concentration units are useful, including percentage, molarity, and molality.
One common way of expressing the relative amount of solute and solvent is through percentage, amount-per-hundred. Percentage can be expressed in three ways:
- mass percent
- mass/volume percent
- volume/volume percent
Mass (Sometimes Called Weight) Percentage
The mass percentage of a solution is the mass of the solute divided by the mass of the solution, multiplied by 100% to get percentage. The mass is commonly measured in grams.
- mass % = (grams of solute/grams solution) × 100%
For example, a solution is prepared by dissolving 25.2 g of sodium chloride in 250.0 g of water. Calculate the mass percent of the solution.
A common error is forgetting to add the solute and solvent masses together in the denominator.
When solutions of this type are prepared, the solute and solvent are weighed out separately and then mixed together to form a solution. The final volume of the solution is unknown.
The mass/volume percent of a solution is the mass of the solute divided by the volume of the solution, multiplied by 100% to yield percentage. The volume of the solution is generally expressed in milliliters.
- mass/volume % = (grams solute/volume of solution) × 100%
When mass/volume solutions are prepared, the grams of the solute are weighed out and dissolved and diluted to the required volume.
For example, a solution is prepared by mixing 125.0 g of benzene with 250.0 g of toluene. The density of benzene is 0.8765 g/mL, and the density of toluene is 0.8669 g/mL. Determine the mass/volume percentage of the solution. Assume that the volumes are additive.
First, determine the volume of the solution.
Notice that it is not necessary to know the chemical formula of either constituent. A common error is forgetting to add the solute and solvent volumes together.
The third case is one in which both the solute and solvent are liquids. The volume percent of the solution is the volume of the solute divided by the volume of the solution, multiplied by 100% to generate the percentage.
- volume % = (volume solute/volume solution) × 100%
When volume percent solutions are prepared, the mL of the solute are diluted with solvent to the required volume.
For example, determine the volume percentage of carbon tetrachloride in a solution prepared by dissolving 100.0 mL of carbon tetrachloride and 100.0 mL of methylene chloride in 750.0 mL of chloroform. Assume the volumes are additive.
A common error is not to add all the volumes together to get the volume of the solution.
If the solute is ethyl alcohol and the solvent is water, then another concentration term is used, proof. The proof of an aqueous ethyl alcohol solution is twice the volume percent. A 45.0 volume % ethyl alcohol solution would be 90.0 proof.
Percentage concentration is common in everyday life (3% hydrogen peroxide, 5% acetic acid, commonly called vinegar, etc.). The concentration unit most commonly used by chemists is molarity. Molarity (M) is the number of moles of solute per liter of solution.
- M = moles solute/liter solution
In preparing a molar solution, the correct number of moles of solute (commonly converted to grams using the molar mass) is dissolved and diluted to the required volume.
Determine the molarity of sodium sulfate in a solution produced by dissolving 15.2 g of Na2SO4 in sufficient water to produce 750.0 mL of solution.
The most common error is not being careful with the units. Grams must be converted to moles, and milliliters must be converted to liters.
Another way to prepare a molar solution is by dilution of a more concentrated solution to a more dilute one by adding solvent. The following equation can be used:
- (Mbefore)(Vbefore) = (Mafter)(Vafter)
In the preceding equation, before refers to before dilution and after refers to after dilution.
Let's see how to apply this relationship. Determine the final concentration when 500.0 mL of water is added to 400.0 mL of a 0.1111 M solution of HC1. Assume the volumes are additive.
The most common error is forgetting to add the two volumes.
Sometimes the varying volumes of a solution's liquid component(s) due to changes in temperature present a problem. Many times volumes are not additive, but mass is additive. The chemist then resorts to defining concentration in terms of the molality. Molality (m) is defined as the moles of solute per kilogram of solvent.
- m = moles solute/kilograms solvent
Notice that this equation uses kilograms of solvent, not solution. The other concentration units use mass or volume of the entire solution. Molal solutions use only the mass of the solvent. For dilute aqueous solutions, the molarity and the molality will be close to the same numerical value.
For example, ethylene glycol (C2H6O2) is used in antifreeze. Determine the molality of ethylene glycol in a solution prepared by adding 62.1 g of ethylene glycol to 100.0 g of water.
The most common error is to use the total grams in the denominator instead of just the grams of solvent.
Practice problems for these concepts can be found at:
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