Lesson Title: Making Solutions is Simple
Objective: This lesson will focus on concepts in multiplication, division and will orient students to simple concepts in algebra and chemistry and biology
Materials:
Graduated cylinders 100 ml
Food coloring
Sugar
Salt
Distilled Water
Workup
Solution chemistry is simple. For instance, say I have 100 ml of water and I add 10 drops of blue food coloring dye to it. I hold then a solution with 10 drops of dye (dod) / 100 ml water (10dod / 100 ml water). Next I pour 50 ml of that solution into another 100 ml graduated cylinder and add another 50 ml of water. Now I have a total volume again of 100 ml but I only have 5 drops of dye in it. So my final concentration is 5 dod / 100 ml.
Demonstrate this simple exercise.
Concentrations
The next exercise will involve sugar. Begin with 100 ml of water and add 4 grams of sugar. Now we have a solution of 4 grams of sugar per 100 ml of water (4g sugar/ 100 ml H2O). STRESS to students that this is the same thing as 0.04 g/ml. Next add 60 ml of this solution to a 100 ml cylinder and bring to a final volume of 100 ml. What is the concentration of the final solution? To solve this problem we should us the following equation
Beginning Concentration X Beginning Volume = Ending Concentration X Ending Volume
OR
VBegin x CBegin = VEnd x CEnd
At this point, present this equation on the chalk board.
So we have VBegin = 60 ml and CBegin = 0.04g/ml. We know that VEnd = 100 ml and we nee to figure out what CEnd is. This is simple first multiply 60x0.04=2.4 (g) next simply divide by the VEnd which is 100 ml. So 2.4g / 100ml = 0.024 g/ml
Salinity
Next introduce the concept of salinity. Salinity is measured in parts per thousand or ppt.
The salinity of seawater is typically in the range of about 33-35 parts per thousand (ppt). One part per thousand means that there is one gram of salt per 1000 grams of water. One ml of water weighs one gram. This may seem extremely convenient but this occurrence is no coincidence. The definition of a milliliter and a gram are based on the physical property known as the density of water.
100 ml of water will weigh 100 grams. So if we want to make artificial seawater that is 35 part per thousand (ppt) we will have to add 3.5 grams of our artifical sea salt. If we had 1000 ml (1 L) of water it would weigh 1000 g (1 Kg) and we would have to add 35 grams of our artifical sea salt to get an artificial sea water solution of 35 ppt.
To demonstrate the concept of salinity ppt letŐs begin by making a solution that is 100 ppt.
Fill a 100 ml graduated cylinder with 100 ml of distilled water.
Next weigh 10 grams of salt and add it to the water.
This is a 100 ppt solution and can be measured using the refractometer.
Dilution
Next letŐs prepare another solution of a different concentration by DILUTING some of our 100 ppt solution.
LetŐs take 70 ml of our 100 ppt solution and pour it into another 100ml grad. cylinder.
VBegin = 70 ml and CBegin = 100 ppt
Next letŐs bring the final volume to 100 ml
VEnd = 100 ml
VBegin X CBegin = VEnd X CEnd
7000 (mlppt) =100 x CEnd
7000 / 100= 70 ppt
Measure this solution with the refractometer
Dilution 2
LetŐs try another one. Pour 50 ml of the 70 ppt solution into another 100 ml cylinder.
VBegin = 50 ml and CBegin = 70 ppt
Bring the final volume to 90 ml
VEnd = 90 ml
VBegin X CBegin = VEnd X CEnd
3500 (mlppt) = 90 m x CEnd
3500 / 90 = 39 ppt
Measure this solution with the refractometer