Question

In a chromatography experiment, a solution containing 5.32 mM analyte yielded a peak whose area is twice as large as the peak area for 2.36 mM standard. When 1.00 mL of 6.93 mM standard was added to 5.00 mL of unknown and diluted to 10.0 mL, the solution gave peak areas of 9365 and 5046 for unknown and standard, respectively. Determine the concentration of unknown in the original solution. A. 1.29 mM B. 2.57 mM C. 12.9 mM D. 25.72 mM

          In a chromatography experiment, a solution containing 5.32 mM analyte yielded a peak whose area is
twice as large as the peak area for 2.36 mM standard. When 1.00 mL of 6.93 mM standard was added
to 5.00 mL of unknown and diluted to 10.0 mL, the solution gave peak areas of 9365 and 5046 for
unknown and standard, respectively. Determine the concentration of unknown in the original solution.
A. 1.29 mM
B. 2.57 mM
C. 12.9 mM
D. 25.72 mM

In a chromatography experiment, a solution containing 5.32 mM analyte yielded a peak whose area is
twice as large as the peak area for 2.36 mM standard. When 1.00 mL of 6.93 mM standard was added
to 5.00 mL of unknown and diluted to 10.0 mL, the solution gave peak areas of 9365 and 5046 for
unknown and standard, respectively. Determine the concentration of unknown in the original solution.
A. 1.29 mM
B. 2.57 mM
C. 12.9 mM
D. 25.72 mM

Added by Janice W.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert David Collins

11/25/2023

Step 1

To calculate the concentration of the standard in the diluted solution, we can use the formula: \[C_1V_1 = C_2V_2\] Where: \(C_1\) = concentration of the standard in the original solution \(V_1\) = volume of the standard in the original solution \(C_2\) = Show more…

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