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17 July, 04:08

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?

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  1. 17 July, 04:14
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    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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