Question

1.02 mg of a compound that is known to absorb at 340 nm is dissolved and diluted to 5.00 mL in a volumetric flask. A 1.00 mL aliquot (portion) of this solution was withdrawn from this flask, placed in a 10.00 mL volumetric flask and diluted to the mark. Some of this solution from the 10.00 mL flask was placed in a cuvet and analyzed analyzed spectroscopically. From these measurements it was found that the concentration was 6.97 x 10-5 M. Calculate the concentration of the analyte in the 5-mL flask

Answer 1

Answer:

$6.97 X 10^{-4}$ M

Explanation:

The concentration of the analyte in the 5-mL flask would be $6.97 X 10^{-4}$ M

This is a problem of simple dilution that can be solved using the dilution equation;

C1V1 = C2V2,

where C1 = initial concentration, V1 = initial volume, C2 = final concentration, and V2 = final volume.

In this case, the initial concentration (C1) is not known, the initial volume (V1) is 1.00 mL, the final concentration is 6.97 x 10-5 M, and the final volume is 10.00 mL.

Now, let us make the initial concentration the subject of the formula from the equation above;

C1 = C2V2/V1. Solve for C1 by substituting the other parameters.

C1 = 6.97 x 10-5 x 10/1 = $6.97 X 10^{-4}$ M