Suppose you perform a serial dilution of Amylase from a stock solution with a concentration of 500ug/mL. In five tests tubes you add 4mL of water. Then you aliquot 1mL of the stock solution into test tube 1 and vortex. From there you aliquot 1mL from test tube 1 into test tube 2. You continue to mix this way into each consecutive test tube. What is the final concentration in test-tube 3?

The final concentration in test-tube 3 is = 4ug/mL

Explanation:

Dilution factor = V initial / V final

Concentration of Stock solution = 500 ug/mL

5 test tubes are prepared with 4 mL of water

1 ml of stock solution is added to the 1st test tube.

therefore, the total volume of 1st test tube is = 4 ml + 1 ml = 5ml

∴ the concentration of 1st Test tube →

 C = m/V   [m= mass (solute) , v= volume]

     =500 ug / 5 ml  = 100 ug/ml

From 1st test tube, 1 ml is taken out and that is added to the 2nd  test tube.

∴ the concentration of 2nd Test tube = 100 ug / 5ml

                                                                = 20 ug/ml

From  2nd test tube, 1 ml is taken out and that is added to the 3rdd  test tube.

∴ the concentration of 3rd Test tube = 20 ug / 5ml

                                                              = 4 ug/ml

The final concentration in test-tube 3 is = 4ug/mL

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