- Units & Notation
- Moles per litre
- Grams per litre
- Percent solutions
- Parts per million
- Practice problems
Serial dilution
Activity 2
Suggest a sequence of 1:100 and 1:10 dilutions that could be performed to reduce the concentration of a solution from 30 g/L to less than 1 µg/mL?
Try to attempt this problem yourself. You may wish to follow the steps identified in the example provided in the previous section to solve problems like these. Type-in your answer (the number of 1:100 dilutions followed by the number of 1:10 dilutions, in digits only) in the text box given below and then check your answer by clicking the 'Check Answer' button.
30 g/L = 30 × 106 µg/L = 30 × 103 µg/mL = 30000 µg/mL
Two 1:100 dilutions followed by a 1:10 dilution would give a dilution factor of 100 × 100 × 10 = 100000
So the final concentration would be 30000 ÷ 100000 = 0.3 µg/mL
You can go to another section of this resource if you have got the answer right, or else, the video below shows a detailed step by step approach to solve the above problem.
This video contains sound.
