Internal Standard Drivers who are suspected of DUI are given a breath analyzer t

Question

Internal Standard Drivers who are suspected of DUI are given a breath analyzer test to see if they exceed the legal limit of alcohol. In cases where there is an accident, injury, or death, the analysis is done via analyzing a blood sample by gas chromatography. A 5.00 mL blood sample from a suspected drunk driver is spiked with 0.500 mL of aqueous 0.750 % propanol as an internal standard. A 10.0 µL portion of the mixture is injected into the GC and the peak areas recorded. Standards are treated in an identical fashion. The following results were obtained: % EtOH (wt/vol) peak area EtOH peak area PrOH 0.020 114 457 0.050 278 449 0.100 561 471 0.150 845 453 0.200 1070 447 Blood sample 782 455 (a) Use a spreadsheet to evaluate the data using the internal standard method. (b) Determine the blood alcohol level of the driver. (c) Does she/he exceed the legal limit of 0.08 %

A water sample was analyzed by ion chromatography to determine the concentration of Mg^2+ and gave a peak area of 0.064 mu S-min. Then 5.0 mL of water sample was spiked with 1.00 mL of 20.00 ppm Mg^2+ and diluted to mark with deionized water. The resulting solution gave a peak area of 0.09 mu S-min. What is the concentration of Mg^2+ in the water sample? A 3.12 stick of chewing gum is asked at high temperature, dissolved in 5 mL of concentrated nitric acid and then diluted to the mark in a 100 mL volumetric flask. To 25 mL volumetric flasks are added 5.00 mL aliquots of the dissolved chewing gum sample and an additional volume of 300.0 ppm aluminum. After dilute to volume, the solutions are analyzed for aluminum by graphite furnace atomic absorption spectrometry at 309 nm. The data are shown in the following table:

Explanation / Answer

1.

The increase in the peak area = 0.095 -0.064 = 0.031 is due to the contribution from 1.00 mL of 20.00 ppm of Mg2+.

So, a peak area of 0.031 is given by 1.00 mL of 20.00 ppm of Mg2+.

therefore, a peak area of 0.064 is given by 1.00 mL of 20.00 x 0.064/ 0.031 = 41.30 ppm of Mg2+.

But the sample was 5.00 mL, hence the concentration of Mg in sample = 41.30/ 5 = 8.26 ppm of Mg2+.

2.

(a) The average increase in absorbance for 0.10 mL of 300.0 ppm Al solution = 0.172 + 0.168 + 0.174 / 3 = 0.171

So, absorbance of 0.171 is found for 0.10 mL of 300.0 ppm of Al

therefore an absorbance of 0.214 will be for 0.10 mL of 300.00 x 0.214/ 0.171 = 375.44 ppm of Al

But the aliqout was of 5.00 mL, hence the conc. of Al in 5.0 mL = 375.44 x 0.10/ 5 ppm of Al = 7.51 ppm of Al.

(b) Wt. % of Al = 7.51 x100/ 1000000 = 0.000751 %.

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