Question

A. In some cases, multiple-step dilutions are required during sample preparation. Consider a case in which 50μL of concentrated dye solution was first diluted to 250mL with distilled water. From this resulting solution, 600μL was transferred to a 100mL volumetric flask and diluted with water to the mark. Calculate the dilution multiplication factor that will be needed to be applied before reporting the concentration in the original solution. Show your calculations. B. Occasionally, the volume of a sample must be reduced so as to increase the concentration of a specific analyte. Consider a situation where 25 liters of an aqueous solution was boiled to evaporate the water. The final reduced volume of the resulting solution was 600mL. Calculate the “dilution multiplication factor” to be used and what fold concentration does this represent? Show your calculations. C. During the recording of her data, a student correctly lists the data for the standard curves in the appropriate columns, however, she accidentally switches the absorbance readings of her unknown so that the 630nm and the 500nm data are listed in the wrong columns. Describe how this will affect her final report for the concentration of the red dye?

Answer 1

Answer A.

50x10^-3 mL of the concentrated dye diluted with 250 mL water, so the total volume = (250+0.050) mL = 250.05 mL

so for the first dilution, the new concentration , say,S1 = 0.050/250.050 = 0.00019996 = 199.96 ppm (ppm = parts per million)

For the secon dilution,

We can apply V1S1 = V2S2, where the V's are corresponding volumes and S' are corresponding concentration.

Here V1 = 650 micro liter = 650x10^-3 mL = 0.650 mL

S1 = 199.96 ppm

V2 = 100 mL,

S2 = to be determined.

By replacing the values,

S2=V1S1/V2 = 0.065*199.96/100 = 0.129974 ppm

Answer B

25000 ml of solution is concentrated to 600 ml of residue.

So, the dilution multiplication factor would be =600/25000 = 0.024

It represent, C =1/0.024 = 41.67 fold concentration.

Answer C

More information required.