BIOL 1406
PreLab 2.3
How can I prepare solutions with specific solute concentrations when I have separate solutes and solvent?
The first method that we will examine for preparing solutions is used when you are starting with separate solutes and solvent. In other words, there is no existing stock solution. In this case, you dissolve the solutes (usually solid) in the solvent (usually dH20.) To make a solution in this way, you must know how much of each solute to use and how much solvent to use. As mentioned previously, a 1 molar (1 M) solution contains 1 mole of solute dissolved in 1 liter of total solution. Obviously, we cannot measure moles directly because we have no way to count atoms or molecules. However, the formula weight of a substance tells us how much one mole of the substance weighs. Therefore, we can use the formula weight to measure moles indirectly.
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Potassium permanganate |
Graduated cylinder with water |
YOUR TURN | ||
You want to prepare one liter of a 2 M potassium permanganate (KMnO4) solution. | ||
How many moles of potassium permanganate should you dissolve in one liter of solution? |
YOUR TURN | ||
The formula weight of KMnO4 is 158.04 g/mole. How much potassium permanganate should you weigh out in order to prepare one liter of the following KMnO4 solutions? | ||
2 M | Hint | Check your answer. |
0.75 M | Hint | Check your answer. |
YOUR TURN | ||
Suppose you do not need an entire liter of solution. How much potassium permanganate should you weigh out in order to prepare the following amounts of a 0.75 M KMnO4 solution? | ||
0.5 L | Hint | Check your answer. |
350 mL | Hint | Check your answer. |
When preparing a solution by dissolving a solid solute in the solvent, use the following formula to calculate the weight of solute needed: Weight of solute (g) = formula weight of solute (g/mole) X molarity (mole/L) X final volume (L) |
For example, to make 600 mL of 0.35 M KMnO4, use the following values:
Formula weight of KMnO4 | = 158.04 g/mole |
Molarity desired | = 0.35 mole/L |
Final volume | = 0.6 liters |
Weight of KMnO4 needed = 158.04 g/mole X 0.35 moles/L X 0.6 L = 33.19 g
(Notice how all the
units except for grams cancelled out of this equation!)
YOUR TURN | ||
Calculate the weight of potassium permanganate (KMnO4 ) needed to prepare 80 mL of a 0.01 M solution. You will need this value during lab. Write the answer in the space provided and also in your lab notebook. |
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Weight of potassium permanganate (KMnO4) needed = |
Once you have weighed out the required amount of solute, place it in
an appropriate volumetric measuring device (such as a graduated cylinder) and
add enough solvent until the total volume of the solution reaches the desired
amount. Notice that if you want to make 80 mL of solution, you must add less
than 80 mL of solvent to the solute because the solute occupies some volume. If
you added a full 80 mL of solvent to the solute, the total volume of the
solution would be greater than 80 mL, and your solution would be too dilute.
General Procedure for Preparing a Solution by Dissolving a Solid Solute in the Solvent:
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YOUR TURN | ||
Your Turn |
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A student wants to make 80 mL of a 1.0 M solution of sucrose (formula weight 342.3 g/mole). How much sucrose should he weigh out to prepare this solution? | Hint | Check your answer. |
The student weighs out the correct amount of sucrose, places it in a beaker, measure 80 mL of dH2O with a graduated cylinder, adds the dH2O to the beaker, and then mixes the solution with a stir bar until all of the sucrose is dissolved. What did he do wrong? | Hint | Check your answer. |
If the final volume of the solution he prepared was 95 mL, what is the actual molarity of his solution? | Hint | Check your answer. |
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