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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 evidence presented in the chromatogram is that the peak corresponding to ethanol at a retention time of 0.9 minutes is clearly distinguishable from other peaks. This implies that ethanol can be identified without interference from the other volatile chemicals present in the sample. The presence of distinct peaks ensures that the qualitative analysis of ethanol remains reliable.
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 is specifically associated with ethanol, which is the primary analyte of interest in this context. Measuring only this peak minimizes the influence of other volatile substances that may not be ethyl alcohol. This results in a more accurate representation of the blood alcohol content, as other peaks may represent different chemicals that could skew the results if included in the analysis.
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 110,000.
Answer
To determine the percentage (m/v) of alcohol, we can use the formula:
ext{Concentration (m/v)} = rac{ ext{Mass of solute (g)}}{ ext{Volume of solution (mL)}} \times 100
Given that the area of the peak corresponding to ethanol is 110,000, the standard curve must be referenced to convert this peak area to mass. Assuming the correlation from the standard curve gives a mass of 0.110 g of ethanol in 1 mL of solution, we have:
ext{Concentration (m/v)} = rac{0.110 ext{ g}}{1 ext{ mL}} \times 100 = 11 ext{ % (m/v)}
Thus, the blood alcohol content would be 11 % (m/v).