Answer:
0.0136mg DDT / g spinach
Explanation:
Quantification in chromatography by internal standard has as formula:
RF = Aanalyte×Cstd / Astd×Canalyte <em>(1)</em>
<em>Where RF is response factor, A is area and C is concentration</em>
Replacing with first experiment values:
RF = 5019×3.20mg/L / 8179×6.37mg/L
RF = 0.308
In the next experiment, final concentration of chloroform was:
11.45mg/L × (1.25mL / 25.00mL) = <em>0.5725mg/L</em>
From (1), it is possible to write:
Aanalyte×Cstd / Astd×RF = Canalyte
Replacing:
6821×0.5725mg/L / 14061×0.308 = Canalyte
Canalyte = <em>0.9017mg/L</em>
as the sample was made from 0.750mL of extract. Concentration of extract is:
0.9017mg/L × (25.00mL / 0.750mL) = 30.06mg/L. As the extract has a volume of 2.40mL:
30.06mg/L × 2.40x10⁻³L = <em>0.07213mg of DDT in the extract</em>
As the extract was made from 5.29g of spinach:
0.07213mg of DDT in the extract / 5.29g spinach = <em>0.0136mg DDT / g spinach</em>
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