Question

A sample of a concentrated fluorescent reagent was serially diluted with three steps using a variable volume 100 to 500 μL micropipetor and 10.00 mL Class A volumetric flasks. The concentrated reagent had a fluorescent signal of 653.4 and the diluted sample produced 0.057 fluorescence units. Unfortunately the steps followed for the dilution were lost. What were those steps?

Answer 1

Preapration of a standard solution produces 653.4 flurescence unit,

consider the solute 1gm dissolve in 10 ml diluent produces 653.4 flurescence unit,

Now, this solution was diluted three times such a that the flurescence produce 0.057 flurescence unit.

1) Take 500 μL above standard solution and diluted to 10 ml volumetric flask.

therefore the quantity of solute in first dilution can be calculated as,

= weight of solute in solution x volume of sample / total volume of solution prepared

= 1000 x 500 / 10000 = 50 mg in 10 ml. .................(1)

2) Take 500 μL above first diluted solution and diluted to 10 ml volumetric flask.

The quantity of solute in second dilution can be calculated as,

= weight of solute in solution x volume of sample / total volume of solution prepared

= 50 x 500 / 10000 = 2.5 mg in 10 ml. .................(2)

3) Take 350 μL above second diluted solution and diluted to 10 ml volumetric flask.

The quantity of solute in second dilution can be calculated as,

= weight of solute in solution x volume of sample / total volume of solution prepared

= 2.5 x 350 / 10000 = 0.1 mg in 10 ml. .................(3)

# From all of above three diluted solution we got weight of solute dissolve in given volume, Now from this equations we can calculate the change in flurescence of solution.

a) The change in intensity of flurescence of solution = intensity of input solution x weight of solute in solution / weight of solute present in solution taken for dilution

= 653.4 x 50 /1000

The change in intensity of flurescence of solution = 32.67 flurescence unit ................( first dilution)

b) The change in intensity of flurescence of solution = intensity of input solution x weight of solute in solution / weight of solute present in solution taken for dilution

= 32.67 x 2.5 / 50

The change in intensity of flurescence of solution = 1.6335 flurescence unit ................( second dilution)

c) The change in intensity of flurescence of solution = intensity of input solution x weight of solute in solution / weight of solute present in solution taken for dilution

= 1.6335 x 0.0875 / 2.5

The change in intensity of flurescence of solution = 0.057 flurescence unit ................( third dilution)

Therefore,

The solultion was diluted as 500μL, 500μL, 350μL respectively.