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AssignmentTenKey - Take-Home Assignment#10 Answer Key 1...

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Take-Home Assignment #10 – Answer Key 1. Shown below is a table summarizing my measurements for retention times and peak widths; note that these retention times are rounded to the closest 0.1 mm and that peak widths are rough estimates, particularly for peaks that are not fully resolved. To find t m , I measured the distance between 8.0 min and 20.0 min as 131.5 mm; thus, the mobile phase elution time of 45 sec corresponds to 8.22 mm. Your calculated results for k, H and N likely will differ slightly from these. Plate heights are in mm. Peak tr (mm) w (mm) k N H (mm) 1 100.7 4.5 11.247 8007 0.0187 2 118.2 6.0 13.376 6206 0.0242 3 137.2 7.0 15.687 6144 0.0244 4 141.2 7.0 16.174 6507 0.0231 5 147.7 7.0 16.965 7120 0.0211 6 173.2 7.0 20.067 9792 0.0153 7 182.7 7.5 21.223 9491 0.0158 8 189.2 8.0 22.013 8946 0.0168 9 217.7 9.0 25.481 9359 0.0160 Average 7953 0.0189 2. A flow rate of 1.2 mL/min and an elution time of 45 sec for a non-retained solute gives the volume of mobile phase as 0.9 mL, which we take as V min . The maximum volume is for a peak eluting at 30 min, which at the stated flow rate is 36 mL. The peak capacity, therefore, is 83 9 . 0 0
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