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- You are preparing a 100.0 mL standard solution needed for titration analysis. The NaOH (40 g/mol) pellets are measured to be 0.5341 g in a balance with acceptable uncertainty of +0.001 g. Mass of the pellets is recorded after taring the container. The pellets are put in a beaker and added with 50 ml distilled water measured by a graduated cylinder. The solution is then quantitatively transferred to a 100-mL volumetric flask with uncertainty of ±0.08 mL and diluted to a 100-mL mark. What is the concentration (M) of the solution and estimate its uncertainty by propagation? Assumption: There is no uncertainty in molar mass. Express your answer as C+/- u M.A student determined fluoride in a toothpaste sample using ion chromatography. A toothpaste sample of mass 0.100 g was dissolved in type I water and then quantitatively filtered through a Whatman #1 filter paper into a clean 100 ml volumetric flask. The flask was then made up to volume with type I water and mixed thoroughly.A portion of the solution was filtered through a 0.45 um filter into a sample vial for analysis. Analysis of the sample gave a fluoride concentration of 1.60 mg/l Calculate the % w/w fluoride in the original toothpaste sample. (Give your answer to 3 decimal places please)A 2.0 g of MgCO3 was dissolved and diluted to exactly 500-mL volumetric flask. If 50-mL aliquot was used in the analysis, what is the weight of MgCO3 present in the aliquot sample
- B. Determination of Total Acid Content in Commercial Vinegar Sample Brand of Sample: Trial 1 Trial 2 Trial 3 Volume of sample 10.00 mL Final reading - standard NaOH solution 25.63 mL Initial reading - standard NaOH 0.08 mL solution Volume consumed - standard 25.55 mL NaOH solution 0.3059 M Molarity of standard NaOH solution Total Acidity in Sample 4.69% Mean Total Acidity in Sample 4.69% BALANCED CHEMICAL EQUATION/S INVOLVED: COMPUTATIONS:In the Analytical Chemistry laboratory, one of the B.Sc student Ms. Fatma wanted to analyze the presence of Iron(II) ion in the water samples collected from Sohar Industrial Area in the Sultanate of Oman and analyzed by Spectrophotometric method. (1) 250.00 mL of this water sample (Solution A) known to contain unknown amount of FeSO4. She has diluted Solution A by a dilution factor 10 to make 250.00 mL solution, which she labeled Solution B. Using Spectrophotometer, she has measured the absorbance value (0.642) for Solution B at 508 nm using 1.00 cm cell (cuvet). The molar absorptivity value for Fe2* ion at 508 nm is ɛ508 = 30.8 M-1cm1. (i) What volume of the solution A did she require to make the 250.00 mL of Solution В? (ii) in (i)? What instrument should she use to transfer the volume of the Solution A calculated (ii) Concentration of FeSO4 in Solution B. (iv) Concentration of FeSO4 in Solution A. Mass of FeSO4 in Solution A (Show your calculation) (v)A 15.00 g sample containing mixed alkali and other inert components was dissolved and diluted to 300 mL with water. A 20 ml, aliquot was titrated with 5.02 mL of 0.5352 M HCI to reach PHP endpoint. Another 20 mL aliquot was titrated to the BCG endpoint, using up 18.87 mL of titrant in the process. Identify the alkali components and their percent weight.
- The standard addition method is used to analyze a sample of a river water for mercury. Solution A is made by pipetting 5.00 mL of undiluted sample in to a 10 mL volumetric flask and filling to the mark with DI water. Solution B is made by pipetting 5.00 mL of undiluted sample and 3.00 mL of 15.0 ppb of Hg standard into same 10.0 mL volumetric flask and filling to the mark with DI. Solution A and B are analyzed using atomic absorption spectroscopy and give a percent transmittance values of 56 % and 33 % respectively (not blank corrected). A blank has a transmittance of 96%. What is the corrected absorbance of both solution A and B? OA. Solution A: 0.123 Solution B: 0.463 O B. Solution A: 0.463 Solution B: 0.234 O C. Solution A: 0.123 Solution B: 0.234 O D. Solution A: 0.234 Solution B: 0.463The standard addition method is used to analyze a sample of a river water for mercury. Solution A is made by pipetting 5.00 mL of undiluted sample in to a 10 mL volumetric flask and filling to the mark with DI water. Solution B is made by pipetting 5.00 mL of undiluted sample and 3.00 mL of 15.0 ppb of Hg standard into same 10.0 mL volumetric flask and filling to the mark with DI. Solution A and B are analyzed using atomic absorption spectroscopy and give a percent transmittance values of 56 % and 33 % respectively (not blank corrected). A blank has a transmittance of 96%. What is the corrected absorbance of both solution A and B? A. Solution A: 0.123 Solution B: 0.463 B. Solution A: 0.463 Solution B: 0.234 C. Solution A: 0.123 Solution B: 0.234 D. Solution A: 0.234 Solution B: 0.463The standard addition method is used to analyze a sample of a river water for mercury. Solution A is made by pipetting 5.00 mL of undiluted sample in to a 10 mL volumetric flask and filling to the mark with DI water. Solution B is made by pipetting 5.00 mL of undiluted sample and 3.00 mL of 15.0 ppb of Hg standard into same 10.0 mL volumetric flask and filling to the mark with DI. Solution A and B are analyzed using atomic absorption spectroscopy and give a percent transmittance values of 56 % and 33 % respectively (not blank corrected). A blank has a transmittance of 96%. What is the corrected absorbance of both solution A and B? O A. Solution A: 0.123 Solution B: 0.463 O B. Solution A: 0.463 Solution B: 0.234 O C. Solution A: 0.123 O D. Solution A: 0.234 Solution B: 0.234 Solution B: 0.463
- 0.3414 grams of sample (crushed aspirin tablet) was weighed and dissolved in 25 mL ethanol. This solution was transferred into a 100 mL volumetric flask and filled to the mark with deionized water. 20 mL of the solution was filtered through a 0.22 μm pore filter and 503 μL of the filtered sample solution was diluted so, that the final volume of the solution was 12 mL. This sample solution and calibration solutions with known concentration of acetyl salicylic acid (ASA) were injected (injection volume 20 μL) into a HPLC and the following peak areas were measured (expressed as averages from three separate chromatographic runs): Solution Cal 1 Cal 2 Cal 3 Cal 4 Cal 5 Sample Structure of ASA: C(ASA) (mg/L) 0 OH B 53.5 105.9 140.9 195.4 247.4 Average weight of the tablet was estimated as 591.9 mg. What is the mass of ASA in one tablet? Please give the answer with 4 significant digits. Be sure you present the result with units! Peak area (mA.s) 4995 9953 12941 17271 22179 10467.3Precipitimetry The chloride in a 4.321-g food sample was precipitated through the addition of 50.00 mL of a standard AgNO3 solution (10.00 mL AgNO3 = 11.22 mL KSCN). The precipitate was coated with nitrobenzene and the mixture was diluted to 250.0 mL. A 50.00 mL aliquot was taken from the diluted solution and required 4.56 mL back titration with a %3D standard KSCN solution (22.33 mL KSCN = 0.9758 g AgNO3). %3D Formula Masses: AgNO3 = 169.87; CI = 35.45 Calculate the following: 1. Molar concentration of KSCN solution M %3D 2. Molar concentration of AgNO3 solution M %3D 3. % (w/w) chloride in the original sample =ASSAY Calibration solutions Calibration solutions of naproxen in the range 5 – 25 μg/mL were prepared. Sample preparation 20 tablets weighing 12.3819 g were crushed to a fine powder. A portion of the powder (145.4 mg) was shaken with approximately 150 mL of acetic acid (0.05 M) for 5 min and then made up to volume in a 250 mL volumetric flask (stock solution). Approximately 50 mL of the stock solution was filtered and a 25 mL aliquot was diluted to 100 mL in a volumetric flask. 10 mL of the resulting solution was further diluted to 100 mL with acetic acid (0.05 M). Analysis The standards and sample solutions were analyzed by HPLC under the following conditions: Column: octadecylsilyl (ODS), 4.6 mmx150 mm, mobile phase: acetonitrile : 0.05 M acetic acid (85:15), flow rate: 1 ml/min, UV detection at 243 nm. Results: A calibration curve of concentration versus peak area was constructed for the standard solutions and gave the straight-line equation: y = 3555.6x + 85, r = 0.9999 The area…