A student performed a serial dilution involving a 1.00 mol dm–3 solution of NaCl. To do this, the student took 10 cm3 of that solution, added it to a 100 cm3 volumetric flask and diluted with water up to 100 cm3 to create dilution #1. The student then took 10 cm3 of dilution #1, added it to a 100 cm3 volumetric flask and diluted with water up to 100 cm3 to create dilution #2. The student repeated this process 3 more times to get dilution #5.Calculate the concentration of dilution #5 in g cm–3.
Question
A student performed a serial dilution involving a 1.00 mol dm–3 solution of NaCl. To do this, the student took 10 cm3 of that solution, added it to a 100 cm3 volumetric flask and diluted with water up to 100 cm3 to create dilution #1. The student then took 10 cm3 of dilution #1, added it to a 100 cm3 volumetric flask and diluted with water up to 100 cm3 to create dilution #2. The student repeated this process 3 more times to get dilution #5.Calculate the concentration of dilution #5 in g cm–3.
Solution
To solve this problem, we need to understand that a serial dilution is a stepwise dilution of a substance in solution. In this case, the student is diluting the original solution by a factor of 10 each time.
Step 1: Calculate the concentration of the original solution in g/cm3
We know that the molar mass of NaCl is approximately 58.44 g/mol. Therefore, a 1.00 mol/dm3 solution of NaCl is equivalent to a 58.44 g/dm3 solution. Since 1 dm3 = 1000 cm3, this is equivalent to 0.05844 g/cm3.
Step 2: Calculate the concentration of each dilution
Each dilution is 10 times less concentrated than the previous one. Therefore, we can calculate the concentration of each dilution as follows:
- Dilution #1: 0.05844 g/cm3 / 10 = 0.005844 g/cm3
- Dilution #2: 0.005844 g/cm3 / 10 = 0.0005844 g/cm3
- Dilution #3: 0.0005844 g/cm3 / 10 = 0.00005844 g/cm3
- Dilution #4: 0.00005844 g/cm3 / 10 = 0.000005844 g/cm3
- Dilution #5: 0.000005844 g/cm3 / 10 = 0.0000005844 g/cm3
So, the concentration of dilution #5 is 0.0000005844 g/cm3.
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