A 500. µl aliquot of a caffeine unknown is diluted to a total volume of 10.00 mL. A calibration curve for caffeine was generated using external standards on the HPLC. The equation of the curve Caffeine Area(mV cm)=1.266C(ppm)+0.029. If the HPLC peak for the diluted unknown has an area of 3.07 mV cm, what is the original caffeine concentration in ppm?
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- percentage purity of an ibuprofen sample was determined by a reversed phase HPLC method as follows: Y Der ● A calibration curve of peak area versus concentration (ug/mL) of pure ibuprofen was constructed The ibuprofen sample for analysis was prepared by taking 10.0204 g of powder and dissolving it in/1Lbf distilled water (solution A). 1 ml of the latter solution was further diluted to IL with distilled water (solution B). Solution B was analysed by HPLC analysis Results The equation for the best-fit line of the calibration curve is: y=26820x-1323.8. ● The average peak area for the ibuprofen sample (solution B) was 266796.2 Calculate the percentage purity of the ibuprofen powder.3c. A solution containing 2.50 mM X (analyte) and 1.20 mM S (standard) gave peak areas of 2509 and 7132 respectively, in a chromatographic analysis. Then, 1.00 mL of 6.00 mM S was added to 5.00 mL of unknown X and the mixture was diluted to 10.0 mL. This solution gave peak areas of 4016 and 3138 for X and S, respectively. Calculate i. the response factor for the analyte ii. the concentration of S (mM) in the 10.0 mL of mixed solutioniii. the concentration of X (mM) in the 10.0 mL of mixed solutioniv. the concentration of X in the original unknown solutionSpectrophotometric Determination of a Metal A 100.00 mL of exactly 250.0 mg/L Cu2+ stock solution was prepared using a 99.95% pure Cu(NO3)2-3H20 (241.60 g/mol). From this stock, a series of standard solutions were prepared by pipetting certain volumes into 100-mL volumetric flasks, followed by adding 2.00 mL of concentrated NH3, and finally diluting to mark with deionized water. Additionally, a blank solution was prepared by using only NH3 and distilled water. Question: Why was NH3 added to each standard solution before diluting with distilled water? What will be the resulting color of the solution and why is this the color that is observed?
- Internal standard (As , Cs ), analyte with known concentration (Ax , Cx ) In a chromatographic equipment, a solution containing 0.006727 X and 0.008331 M S give peak area of Ax= 2316 and As= 207. To analyze an unknown sample, 5.0mL of 0.008331M S was added to 5.0mL of X, and the mixture was diluted to 50.0mL. This mixture gave a chromatography spectrum with area Ax= 3206 and As= 200. Find Cx.Q1: The following are relative peak areas for chromatograms of standard solutions of methyl vinyl ketone (MVK) MVK concentration mmol/L 0.5 Relative peak area 3.76 1.5 9.16 2.5 15.03 3.5 20.42 4.5 25.33 5.5 31.97 A sample containing MVK yielded relative peak area of 21.3. Calculate the concentration of MVK in the sample.(use excel)The following data are from the HPLC assay of paracetamol 500 mg tablets using the calibration curve method: - Weight of active ingredient in powdered tablets taken for analysis: 180 mg • Initial extraction volume: 200 mL - Dilution steps: 10 mL into 100 mL, then 5 mL into 100 mL - Linear equation: y = 12000x - 50 %3! - Area of chromatographic peak for sample is 4750. What is the theoretical concentration in final solution in mg/100 mL? (2 decimal places). Please do not write the units, just numerical value. Answer:
- A 500.0 μL aliquot of a ferric chloride solution is added to a 50.00 mL volumetric flask. To this flask is added 10 mL of 2M sodium acetate, 2 mL of 10 wt% NH4OCI, 3 mL of 5 wt% 1,10-phenanthroline, and enough DI water to fill the remaining volume of the flask for a total volume of 50.00 mL. Based on the UV-Visible light absorbance, the concentration of [Fe2+]=2.6 ppm. What is the concentration of Fe3+ ion in the original ferric chloride solution? O 3.85 X 103 ppm O 260.0 ppm O 260 ppm O 130 ppmA lab pair’s Beer’s Law calibration graph 0.08 < A < 0.79 had a best-fit line of A = 0.013 C – 0.023. Data was: off-scale for both the stock solution and solution #1 (10.00 mL stock diluted to 25.00 mL); solution #2 (1.00 mL stock diluted to 25.00 mL) had A = 0.85, and solution #3 (1.00 mL of solution #1 diluted to 25.00 mL) had A = 0.73. What are the concentrations of solution #3 and the stock solution?A solvent passes through a chromatography column in 3.87 min, but the solute requires 7.60 min. What is the retention factor, k? k = What fraction of the time does the solute spend in the mobile phase in the column? t fraction in mobile phase = The volume of the stationary phase is 0.133 times the volume of the mobile phase in the column (Vs = 0.133VM). What is the distribution constant, Kp, for this system? KD = %3D
- 4.Consider the peaks for pentafluorobenzene and benzene in the gas chromatogram shown here. The elution time for unretained solute is 1.06 min. The open tubular column is 30.0 m in length and 0.530 mm in diameter, with a layer of stationary phase 3.0 μm thick on the inner wall. a) Measruing the width, w, at the baseline on the chromatogram, find the number of plates for these two compounds b) Use your answer to (a) to find the resolution between the two peaks c) Using the number of plates N=sqrtN1*N2 with the values from (a) calcuate what the resolution should be and compare your answer with the measured resolution in bA scientist wishes to measure the concentration of methyl benzoate in a plant stream by gas chromatography. He prepares a sample of butyl benzoate to use as an internal standard. The ic results of a preliminary run, which used a solution known to contain 1.37 mg/mL of methyl benzoate (peak A) and 1.51 mg/mL of butyl benzoate (peak B), are shown. The area of peak A is determined to be 301 and the area of peak B is determined to be 349 measured in arbitrary units by the computer. To measure the sample, 1.00 mL of a standard sample of butyl benzoate containing 2.41 mg/mL is mixed with 1.00 mL of the plant stream material. Analysis of the mixture gave a 10 Time (min) 15 peak area of 499 for peak A and 417 for peak B. What is the concentration of methyl benzoate in the plant stream? methyl benzoate concentration: mg/mL Detector ResponseSuppose a difference in changes in plasma oxytocin levels between two groups of 0.2 pg/mL would be of interest to researchers and that changes within each group have standard deviation o = 0.1 pg/mL. The power of a one-sided two-sample t-test to detect a difference in changes in oxytocin levels of 0.2 pg/mL between two groups of 5 women at the 5% level is