A. In some cases, multiple-step dilutions are required during sample preparation. Consider a case in which 50μL of concentrated dye solution was first diluted to 250mL with distilled water. From this resulting solution, 600μL was transferred to a 100mL volumetric flask and diluted with water to the mark. Calculate the dilution multiplication factor that will be needed to be applied before reporting the concentration in the original solution. Show your calculations.
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Answer A.
50x10-3 mL of the concentrated dye diluted with 250 mL water, so the total volume = (250+0.050) mL = 250.05 mL
so for the first dilution, the new concentration , say,S1 = 0.050/250.050 = 0.00019996 = 199.96 ppm (ppm = parts per million)
For the secon dilution,
We can apply V1S1 = V2S2, where the V's are corresponding volumes and S' are corresponding concentration.
Here V1 = 650 micro liter = 650x10-3 mL = 0.650 mL
S1 = 199.96 ppm
V2 = 100 mL,
S2 = to be determined.
By replacing the values,
S2=V1S1/V2 = 0.065*199.96/100 = 0.129974 ppm
Answer B
25000 ml of solution is concentrated to 600 ml of residue.
So, the dilution multiplication factor would be =600/25000 = 0.024
It represent, C =1/0.024 = 41.67 fold concentration.
Answer C
More information required.
A. In some cases, multiple-step dilutions are required during sample preparation. Consider a case in which...
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