Question:
A chemist was performing a spectroscopy experiment on a fluorescent dye to examine the influence of concentration on absorbance. To accomplish this, the chemist created a stock solution of a certain concentration of dye using 10 mL of water ($4373_w45_h15.png$). Then, a serial dilution was performed by taking 1 mL of the stock solution and diluting it with 9 mL of water for a total solution volume of 10 mL. 1 mL was taken from Solution 2 and diluted with 9 mL water to create Solution 3. The serial dilution was performed through Solution 5, as shown in the diagram below. If the dye concentration in Solution 5 was 0.5 g/mL, what was the concentration of the dye in Solution 2?
A serial dilution is a method used to dilute the concentration of substances. A stock solution is created at a known concentration, and then a certain volume is extracted and diluted with a solvent. Then, the same volume of the next solution is extracted and diluted with the same volume of solvent. The process is continued until the desired number of samples is created. They are useful for creating very dilute solutions, where it would be very difficult to weigh out the amount of reagent.
In this scenario, the concentration of the original stock solution is not provided. You are told that the concentration of Solution #5 is 0.5g/ml. Since we are told that each dilution is 1mL of the solution mixed with 9mL of the solvent, you can work backwards. Each time the solution is diluted, the concentration is reduced by 10 fold (divided by 10). We know this because 1mL of the solution is put until a new solution of 10mL. Therefore, with each dilution, you would divide the amount of the solute by 10.
Since we are working in the opposite direction, we are going to multiply the concentration as we move backwards.
0.5 × 10 = 5 g/mL (Solution 4),
5 × 10 = 50 g/mL (Solution 3)
5 × 100 = 500 g/mL (Solution 2)