A mixture containing 188 m of the analyte of interest and, Ask an Expert

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A mixture containing 18.8 M of the analyte of interest and 44.4 M of the internal standard formed chromatographic peak areas of 306 for the analyte and 511 for the standard. A second solution containing an unidentified quantity of the analyte and 55.5 M of the internal standard produced peak areas of 251 for the analyte as well as 563 for the internal standard.

Explain what is the concentration of the analyte in the second solution expressed as M?

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Category:- Chemistry
Reference No.:- M91224835

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