Hallo,
I had a lignin solution and inorder to mesure the concentration of lignin in the solution i had taken 2000µL of the solution and added 5000µL of DI water to dilute it as the absorbance was more than 1.5 . Can anyone please tell me what the resulting dilution factor is ? I usually consider 2000µL as dilution factor 1 for the measurements.
Thank you in advance :)
Hello, Specifically to determine the dilution factor in this case, you first had 2000 µL of your lignin solution in the tube. Then, you further added 5000 µL of DI water to wash or dilute the solution. This makes you total final volume up to 7,000 µL which comprises of 2000 µL and 5000 µL respectively. It is calculated by the use of the ratio with the total final volume on the numerator and the original volume of the solution in the denominator. In this case: Dilution factor or the dilution rate that is the ratio figuring out how many times a given stock is diluted is calculated thus: Dilution Factor = Final Volume/Initial Volume = 7000/2000 = 3.5 Consequently, the dilution factor is 3.5 so your solution is less concentrated 3.5 times to the first factor. Hope this clears up things! 🙂
After adding 5000µL of DI water to 2000µL of lignin solution, the resulting dilution factor is 3.5. This indicates that the diluted solution is 3.5 times more diluted than the original solution.
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