An HPLC analysis is carried out on codeine phosphate linctus. The linctus is diluted to 50.00 mL with water prior to analysis. The weight of the linctus analyzed was 13.5000 g. The density of the linctus was 1.2522 g/mL using a densitometer. Area of the 0.074400% w/v codeine phosphate standard was 85,585. The area of the codeine phosphate peak obtained from the linctus was 62,880. Calculate the percent w/v of codeine phosphate in the linctus.
The answer is supposed to be 0.2535 but I'm not sure how the professor got that and also, the other answer to this question posted on Chegg is incorrect. Please so all work and thanks in advance!
Concentration is directly proportional to area.
=> (A1/C1) (standard) = A2/C2 (linctus)
=> 85585 / 0.074400% w/v = 62880 / C2
=> C2 = (62880 / 85585) x 0.074400
=> C2 = 0.054660% w/v
0.054660% w/v is the concentration of the diluted linctus
Volume of the original concentrated linctus = mass / density = 13.5000 g / 1.2522 g/mL = 10.78 mL
Hence concentration of linctus in the original concentrated linctus
= 0.054660% w/v x (50.00 / 10.78) = 0.2535% w.v (answer)
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