Question

Practice problem F5: Design a serial dilution that has at least three steps and ends with a DF of 10. The smallest volume that can be transferred between any two tubes is 0.01 ml and the largest volume of diluent in any individual tube is 9.99 L.

Answer 1

Given conditions-

Serial dilution has at least 3 steps, total dilution factor (TDF) of end tube is 10^-8, smallest transferred volume is 0.01 mL, largest amount of diluent is 9.99 mL.

Now in serial dilution, dilution factor will be constant.

Also, dilution factor = Volume of stock solution taken/ Final volume

Now, if we take 0.1 mL of stock solution & if it is transferred to the first tube which contains 9.9 mL of diluent, then dilution factor of first tube = (0.1) / (0.1+9.9) = 0.1/10 = 1/10 = 10^-2 dilution

Now, if this step is repeated 4 times, then TDF of end tube will be = 10^-2 x 10^-2 x 10^-2 x 10^-2 = 10^-8 dilution

Thus, if we transferred 0.1 mL stock solution to 9.9 mL of diluent & repeat this step 4 times, then we will get desired result.

0.1 mL 0.1 ml 0.1 ml 0.1 ml Stock solution 9.9 mL diluent 9.9 mL diluent 9.9 ml diluent 9.9 mL diluent Stock solution 10-2 dilution 10+ dilution 10- dilution 10- dilution