TRIAL 1 TRIAL 2 TRIAL 3 Mass of Vitamin C (mg) 0.011 g/ 11 mg 0.012 g/12 mg 0.011g/ 11 mg Initial buret reading of Iodine titrant, mL 100 ml 100 ml 100 ml Final buret reading of Iodine titrant, mL Volume of Iodine solution used, mL 41.7 ml 38.2 ml 43.9 ml 58.3 ml 61.8 ml 56.1 ml Mass Vitamin C, mg /Volume of Iodine 0.19 mg/ml 0.19 mg/ml 0.20 mg/ ml solution used, mL Average mass of Vitamin C, mg/ volume 0.19 mg/ml of Iodine solution used, mL Molarity 12 Average Molarity 12

TRIAL 1 TRIAL 2 TRIAL 3 Mass of Vitamin C (mg) 0.011 g/ 11 mg 0.012 g/12 mg 0.011g/ 11 mg Initial buret reading of Iodine titrant, mL 100 ml 100 ml 100 ml Final buret reading of Iodine titrant, mL Volume of Iodine solution used, mL 41.7 ml 38.2 ml 43.9 ml 58.3 ml 61.8 ml 56.1 ml Mass Vitamin C, mg /Volume of Iodine 0.19 mg/ml 0.19 mg/ml 0.20 mg/ ml solution used, mL Average mass of Vitamin C, mg/ volume 0.19 mg/ml of Iodine solution used, mL Molarity 12 Average Molarity 12

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter7: Statistical Data Treatment And Evaluation

Section: Chapter Questions

Problem 7.25QAP

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Trial 1 Trial 2 Trial 3 Initial burette reading (mL) 2.29 1.41 1.95 Molarity of NaOH (M) 0.100 0.100 0.100 Volume of vinegar sample (mL) 5.00 5.00 5.00 Final burette reading (mL) 50.37 49.39 49.84 Table 2. Titration data Trial 1 Trial 2 Trial 3 Initial burette reading (mL) 2.29 1.41 1.95 Molarity of NaOH (M) 0.100 0.100 0.100 Volume of vinegar sample (mL) 5.00 5.00 5.00 Final burette reading (mL) 50.37 49.39 49.84 Expected color at end point Volume of NaOH used (mL) 48.08 47.98 47.89 Compute for the ff: a. Average moles of acetic acid (mol)? b. Average molarity of acetic acid (M)? c. Average molarity of acetic acid (M)?
(a) During the analysis of water sample by argentometric titration, results obtained are as follows: Experiment1 2 3 4 Volume 15.5, 15.2 ,15.1 ,15.4 A)Calculate average deviation and relative average deviation for the given data. (b)Calculate the molecular weight of an unknown acid if 8.5 g of it is dissolved in 200.0 ml of water and requires 50.0 ml of 1.5 M sodium hydroxide for complete neutralization .
A 50.00 (±0.03) mL portion of an HCl solution required 29.71(±0.03) mL of 0.01963(±0.0030) M Ba(OH)2 to reach an end point with bromocresol green indicator. The molar concentration of the HCl is calculated using the equation below (attached image): a.) Calculate the uncertainty of the result (absolute error). M=0.02333(±?????) M b.) Calculate the coefficient of variation for the result. CV= (Sy/y) x 100%
Density of solution:Trial 1: 1.2 g/mLTrial 2: 1.2 g/mLTrial 3: 1.2 g/mL Average density = 1.2 g/mL What is the relative average deviaion, %?
An unknown sample of Cu2+gave an absorbance of 0.262. Then 1.00 mL of solution containing 100 ppm (ug/mL)Cu2+was mixed with 95.0 mL of the unknown and the mixture was diluted to 100 mL in a volumetric flask with deionized water. The absorbanceof the new solution was 0.500. a.Denoting the initial unknown concentrationas [Cu2+]i,write an expression for the final concentrationafter dilution [Cu2+] b.Find [Cu2+]in the unknown.
Sample AnalysisSource of Water sample:Tap Water Molarity of Na2S2O3 1 2 3 0.0243 0.0245 0.0228 Average Molarity of Na2S2O3 0.0239 Trials 1 2 3 Volume of sample used 100.00 100.00 100.00 Final Volume Reading Na2S2O3 (ml) 6.70 10.00 13.10 Initial Volume Reading Na2S2O3 (ml) 3.40 6.70 10.00 Net VolumeNa2S2O3 used (ml) 3.30 3.30 3.10 mg O2 in sample ppm O2 in sample average ppm O2 in sample
EX 2.Given-> Volume of Na+ = 500 ml Molarity of Na+= 0.0100M Molar mass of Na2CO3 = 105.99 gm/mole Millimole of Na+ = molarity × volume Number of millimole = 0.0100 × 500 = 5 millimole Na2CO3 ---> 2Na+ + CO32- Millimole of Na2CO3 = millimole of Na+/2 Millimole of Na2CO3 =5/2 = 2.5 millimole Mole of Na2CO3 = 2.5 × 10-3mole (1 mole = 10^3 millimole) Weight of Na2CO3 required = mole × molar mass = 2.5 × 10-3 × 105.99 =0.26 gm Hence, 0.26 gm Na2CO3 must dissolve in 500 ml of water.
Answer the following: Maria was given a capsule of multivitamins and she was asked to determine the % by mass (w/w) of ascorbic acid present in the sample. The student analyzed 1.032 g sample using volumetric titration. Use the table as a reference of the data to be used for your solution. Given Choices: A. 66.0 % B. 54.6% C. 65.6% D. 7.30%
Give typed answer not written You send a sample of rice to the lab for the determination of the total arsenic content. The analyst dissolves 525 mg of the rice in 5.0 mL of concentrated acid and then dilutes the solution to a final volume of 20.0 mL. When analyzed by ICP-MS, the solution concentration is found to be 20 µg L-1. What is the concentration of arsenic in the rice (in µg kg-1)?
Figure 1:Experimental results Mass of empty weighing dish,g 2.2522g Mass of weighing dish + CaCl2 · 2H2O,g 6.2540g Mass of CaCl2 · 2H2O, g 4.0018g [Na₂CO₃], mol⋅L−1 0.3330 mol⋅L−1 Volume of the CaCl2 solution used, mL 10.00mL Volume of the Na₂CO₃ solution used, mL 10.00mL Mass of the filter paper + watch glass, g 51.3999g Mass of the filter paper + watch glass + dry product (final), g Weight 1=51.7235g Weight 2=51.7058g Figure 2:Calculated data Mass of dry product, g 0.3059g Moles of CaCl2 used, mol 0.02721mol Moles of Na₂CO₃ used, mol 0.02721mol [CaCl2], mol⋅L−1 0.3330 mol⋅L−1 Limiting reagent Calculated mass of excess reagent remaining in the mixture after reaction, g Theoretical yield, g % yield I need help in my lab. The number of mols of cacl2 and na2co3 gives me the same number. If I need to calculate the limiting reagent, how would I be able to do that if they have…
The thiourea in a 1.455 g sample of organic material was extracted into a dilute sulfuric acid solution and titrated with 37.31 mL of 0.009372 M Hg2+ via reaction: 4(NH2)2CS + Hg2+ →[(NH2)2CS]4 Hg2+ P.S. Answer only the last two letters of the following questions. (Only C and D) a. Is this an example of total analysis technique or concentration technique? Explain. b. Calculate the percent (NH2)2CS ( 76.12 g/mol) in the sample. c. What is classification of the analysis based on the amount of sample and amount of analytes present? Explain. d. If the true value is 10.00%, calculate the absolute and relative error.
A 190.00 mL solution of 0.00300 M AB4 is added to a 340.00 mL solution of 0.00265 M CD5. What is pQsp for AD4? Answer: (11.251)