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A forensic chemist wants to test the accuracy of a gas chromatograph that is to be used for the analysis of blood alcohol content - VCE - SSCE Chemistry - Question 5 - 2010 - Paper 1
Question 5
A forensic chemist wants to test the accuracy of a gas chromatograph that is to be used for the analysis of blood alcohol content. A blood sample may contain a numbe... show full transcript
Worked Solution & Example Answer:A forensic chemist wants to test the accuracy of a gas chromatograph that is to be used for the analysis of blood alcohol content - VCE - SSCE Chemistry - Question 5 - 2010 - Paper 1
Step 1
What evidence is presented in the chromatogram to support this claim?
Answer
The chromatogram shows that the peak corresponding to ethanol is distinct and clearly separated from the peaks of the other volatile chemicals. This suggests that the presence of other substances does not interfere with the detection of ethanol, allowing for accurate qualitative analysis.
Step 2
To determine the percentage of alcohol in a blood sample, only the peak at a retention time of 0.9 minutes is measured. Explain why.
Answer
The peak at a retention time of 0.9 minutes corresponds specifically to ethanol, which is the substance of interest for measuring blood alcohol content. Other peaks represent different substances that may be present but do not pertain to the measurement of ethanol. Thus, measuring only the peak at 0.9 minutes ensures the accuracy of the alcohol content determination.
Step 3
Determine the percentage (m/v) of alcohol in the driver's blood if the peak area at a retention time of 0.9 minutes was found to be 110000.
Answer
To calculate the percentage (m/v) of alcohol, the area under the ethanol peak can be related to alcohol concentration using a calibration curve or formula specific to the chromatographic method used. If the relationship is established, for example, as concentration (mg/mL) ≈ Area / k with k being a constant derived from calibration, you can then compute the concentration. Assuming the relationship gives 0.1 mL of alcohol for every area unit, the concentration would be:
ext{Concentration} = rac{110000}{k}
Depending on the value of k, you will get the m/v percentage. For example, for k = 1000, concentration = 110 mL/L or approximately 0.11% (m/v).