A tank of water was contaminated with a colored dye (D). To determine the concentration of D, an analyst took 5.0-mL solution from the contaminated water as a sample, and measured the absorbance, yielding a value of 0.10. Afterwards, the analyst added 5.0-mL of 0.02 M standard D solution into the sample solution, and measured the absorbance, yielding a value of 0.20. What is the concentration of D in the original sample solution

0.005 M

Explanation:

Given data:

volume of sample solution ( volume of D ) = 5.0 mL

volume of added stock solution ( V1 ) = 5.0 mL

concentration of added stock solution ( N1 )  = 0.02 M

Total volume of concentration  ( V2 )= 10 mL = ( 5.0 mL + 5.0mL)

concentration of Total volume of sample ( C2 ) = 0.01

N2 = ( N1V1 ) / V2

     = ( 0.02 * 5 ) / 10 = 0.01 m

absorbance of sample solution ( A1 ) = 0.10

absorbance of additional sample solution ( A2 ) = 0.20

attached below is the remaining part of the detailed solution

a. a clone will be physically identical to the original, but won’t be genetically identical

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