Question

20. A series of five ½ dilutions was performed. What is the final dilution? If the original concentration was 50, what is the concentration in tube 4? What is the dilution factor for tube 3? (3 points)

Answer 1

This problem is similar to geometric progression where the initial amount is 50 and the common ratio is 0.5 which is the dilution factor

after 5 dilution the concentration will be $\frac{50}{2^5}=\frac{50}{32}$

concentration in 1st tube will be $50\times\frac12$

concentration in 2nd tube will be $50\times\frac12\times\frac12$

concentration in 3rd tube will be $50\times\frac12\times\frac12\times\frac12$

concentration in 4th tube will be $50\times\frac12\times\frac12\times\frac12\times \frac12$