CK-12-Chemistry Intermediate

http://www.ck12.org Chapter 16. Solutions

• molarity = 0.250 M


• volume = 3.00 L


• molar mass of KMnO 4 = 158.04 g/mol

Unknown

• mass of KMnO 4 =? g

Moles of solute is calculated by multiplying molarity by liters. Then, moles is converted to grams.

Step 2: Solve.

0 .250 M KMnO 4 × 3 .00 L solution= 0 .750 mol KMnO 4

0 .750 mol KMnO 4 ×

158 .04 g KMnO 4

1 mol KMnO 4

=119 g KMnO 4

Step 3: Think about your result.

When 119 g of potassium permanganate is dissolved in enough water to make 3.00 L of solution, the molarity is
0.250 M.

Practice Problem

2. What mass of CaCl 2 is needed to make 600. mL of a 0.380 M solution?

Preparing Solutions

If you are attempting to prepare 1.00 L of a 1.00 solution of NaCl, you would obtain 58.44 g of sodium chloride.
However you cannot simply add the sodium chloride to 1.00 L of water. After the solute dissolves, the volume of
the solution will be slightly greater than a liter because the hydrated sodium and chloride ions take up space in the
solution. Instead, a volumetric flask needs to be used. Volumetric flasks come in a variety of sizes (Figure16.7)
and are designed so that a chemist can precisely and accurately prepare a solution of one specific volume.

In other words, you cannot use a 1-liter volumetric flask to make 500 mL of a solution. It can only be used to prepare
1 liter of a solution. The steps to follow when preparing a solution with a 1-liter volumetric flask are outlined and
shown below (Figure16.8).

1. The appropriate mass of solute is weighed out and added to a volumetric flask that has been about half-filled
   with distilled water.


2. The solution is swirled until all of the solute dissolves.


3. More distilled water is carefully added up to the line etched on the neck of the flask.


4. The flask is capped and inverted several times to completely mix the solution.

Dilutions

When additional water is added to an aqueous solution, the concentration of that solution decreases. This is because
the number of moles of the solute does not change, but the total volume of the solution increases. We can set up an
equality between the moles of the solute before the dilution (1) and the moles of the solute after the dilution (2).