Question

Low concentrations of warfarin near the detection limit gave the dimensionless instrument readings: 203.3, 172.3, 160.3, 172.3, 217.3, 158.1, 221.1. 199.7, 132.7, and 203.5. Ten blanks had a mean reading of 56.3. The slope of the calibration curve is 3.55 x 10' M-!. Estimate the signal and concentration detection limits and the lower limit of quantitation for warfarin. signal detection limit: concentration detection limit: lower limit of quantitation:

Answer 1

Given that low concentrations of warfarin near the detection limit za the following dimensionless instrument readings! 203.3.172.3, 100.3, 172.3, 2173 | 158.1, 221.), 1997, 132.7,203.5 Ten blanks had a mean reading of 56-3. Then the slope of the caliberation curve is 3:55X10 ml Then we have to write mean ñ : 203.3 +112.3 +160.3 + 217-3 + 172-3+. 10 Then the mean ñ value we get mean ū. 184.06 The standard denivation is calculate now then we get s= (hi-nje - N-1 whore, ni: 203.3,172-3. 1603, apos na ñ = 184.06. toute N:10 Then the values we can sulistitute 203.3-184-00)2 + (172.3-1840)? +(Ito.3-184.00 24 S: 29.102 =) opproximately 3 = 29.10 2 Signal limits First we have to find out the Signal limit - y blankt 38 where y blank: 56:3, 3: 29.100 Then, substitute the values then we get = 56.3 + 3x29.10 = 56.3+ 87.3 L LLL : 143.6 The signal limit. 143.6

by Concentration detection limit :- Now we have to find out the concentration detection limit whole, 8:23:10 ma 3.55x 109 MO! Then, substituting the values then 3.55*10 - 3x99.10 3.55x109 873 Tiosx1.55 -), 879 3550000000 2 20459 *10-8 M L L concoch detection limit : 2.453x10-8M 1 s) The lower limit dy govom motion for Cuartozines Now We have to Andout the lower limit ck quantation for warfarina is 10 $ Whene S- 29.10 _ mr.sey 10M m: 3:55X 109 M Then, Substituting the values then we get lox29. 0 3.53X10 7 29.10x10 3.35 10 g) 8.1972108m 73550000 000