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A student needs 25.0 mL of 1.61×10 -4 M FD&C Blue Dye No. 2 solution. If the student has 2.00 mM FD&C Blue Dye No. 2 stock solution to work with, how much of the 2.00 mM solution should be added to the 25.0 mL volumetric flask? Hint, 2.00 mM is 0.00200 M.     4.981 mL     0.4981 mL     3.508 mL     12.45 mL     2.008 mL

A student needs 25.0 mL of 1.61×10 -4 M FD&C Blue Dye No. 2 solution. If the student has 2.00 mM FD&C Blue Dye No. 2 stock solution to work with, how much of the 2.00 mM solution should be added to the 25.0 mL volumetric flask? Hint, 2.00 mM is 0.00200 M.     4.981 mL     0.4981 mL     3.508 mL     12.45 mL     2.008 mL

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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QUESTION 5

  1. A student needs 25.0 mL of 1.61×10 -4 M FD&C Blue Dye No. 2 solution. If the student has 2.00 mM FD&C Blue Dye No. 2 stock solution to work with, how much of the 2.00 mM solution should be added to the 25.0 mL volumetric flask? Hint, 2.00 mM is 0.00200 M.
       
    4.981 mL
       
    0.4981 mL
       
    3.508 mL
       
    12.45 mL
       
    2.008 mL
Expert Solution
Step 1

Given: Concentration of stock solution = 2.00 mM = 0.002 M                       (Since 1 M = 1000 mM)

Final volume of solution = 25.0 mL.

And concentration of final solution = 1.61 × 10-4 M = 0.000161 M.

 

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