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A standard solution with a concentration of 12.5 ppm gave a fluorescence emission intensity of 5.227...

A standard solution with a concentration of 12.5 ppm gave a fluorescence emission intensity of 5.227 x 104 counts at 457 nm. An unknown sample was diltued by placing 5 mL of the sample into a 250 mL volumetric flask and diluted to the mark. The resulting solution gave an emission intensity of 3.921 x 104 counts at 457 nm. Determine the concentration of analyte in the unknown sample in ppm.

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Answer #1

yelated Becys laur 3a5 On K PoG tAL 1R5 PPm HUD I 5.RXIO KPo = 5. र२+x\०५ 12 5 aF diluted 5aMble) IR= 3.gI X1o4 3.921 XLa 5R

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