5 Steps to a 5 AP Chemistry 2019

Solutions and Colligative Properties ❮ 183

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.

Answer:

=

++

volume% ×=

(100.0mL carbon tetrachloride)

(100.0 100.0 750.0)mL solution

100% 10.53%

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.

Molarity

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 Na 2 SO 4 in sufficient water to produce 750.0 mL of solution:

molarity=× ×=

15.2 gNaSO

750.0mL

1mol gNaSO

142 gNaSO

1,000 mL

1L

(^2424) 0.143
24

M

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.

Mbefore = 0.1111 M Mafter =?

Vbefore = 400.0 mL Vbefore = (400.0 + 500.0) mL

Mafter = (Mbefore)(Vbefore)/(Vafter) = (0.1111 M )(400.0 mL)/(900.0 mL)

= 0.04938 M

The most common error is forgetting to add the two volumes.

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