Following the monograph procedure, a 724-mg of aspirin (MW-180 g/mol) dissolved in 18.5 ml of cold neutralized alcohol. This solution was then initially titrated with 0.101 N sodium hydroxide solution, then later neutralized with 0.104 N sulfuric acid. What is the percentage purity of sample
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- Following the monograph procedure, a 724-mg of aspirin (MW=180 g/mol) dissolved in 18.5 ml of cold neutralized alcohol. This solution was then initially titrated with 0.101 N sodium hydroxide solution, then later neutralized with 0.104 sulfuric acid. What is the percentage purity of the sample? 1. What is the milliequivalent weight consumed by the acidic titrant? a. 1.5392 g-meq b. 2.0907 g-meq c. 1.4948 g-meq d. 2.1528 g-meq 2. What is the milliequivalent weight consumed by the basic titrant? a. 5.8656 g-meq b. 1.5392 g-meq c. 5.6964 g-meq d. 1.4948 g-meq 3. What is the difference of milliequivalent weight consumed in the reaction? a. -4.1572 g-meq b. 4.3708 g-meq c. 0.5515 g-meq d. 4.1572 g-meq20 tablets containing the dibasic drug ethambutol hydrochloride (MW 277.2 g/mol) were powdered. 0.3354 g of tablet powder was dissolved in 10 mL of 2 M sodium hydroxide and was extracted with 5 x 25 mL quantities of chloroform. The combined extracts were evaporated and the residue was dissolved in 100 mL of anhydrous acetic acid. Non-aqueous titration was carried out using 1-naphtholbenzein solution as indicator and 0.1014 M acetous perchloric acid. Volume of acetous perchloric acid required = 14.2 mL, weight of 20 tablets = 3.367 g, state content of ethambutol hydrochloride per tablet = 100 mg. Calculate the % of stated content in the tablets.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+ Is this an example of total analysis technique or concentration technique? Explain.
- A 0.1093-g sample of impure Na2CO3( Molecular weight 106) was analyzed by the Volhard method. After adding 50.00 mL of 0.06911 M AGN03 (Molecular weight 169.87), the sample was back titrated with 0.05781 M KSCN, requiring 27.36 mL to reach the end point. Report the purity of the Na2CO3 sample.Measure the mass of 0.400g of Co(NO3)2 + 6H2O and dissolve it in 15mL of distilled or deionized water using a clean beaker Then while using a graduated cylinder, measure out 10.0mL of a prepared 0.10-M Na3PO4 soultion. Determine which compound Na3PO4 or Co(NO3)2 *6H2O will be the limliting reagent. Then detrimine the theiretical yeildIn the conductivity test, _____________________ is going to result to a brightly lit light bulb. a. 70% v/v ethanol b. 1.0 M citric acid c. Mixture of 10.0 mL 1.0 M Mg(OH)2 and 10.0 mL 1.0 M HNO3 d. Glacial acetic acid
- A 0.1093-g sample of impure Na2CO3 was analyzed by the Volhard Method. After adding 50.00 mL of 0.06911 M AgNO3, the sample was back titrated with 0.05781 M KSCN, requiring 27.36 mL to reach the end point. Report the purity of Na2CO3 sample.0.56 g 1,4‐dimethoxybenzene, 0.9 mL 2‐chloro‐2‐methylpropane, 5 mL nitrobenzene and 0.52 g graphite were combined, and the mixture was refluxed for 1.5 hours. It was then cooled to room temperature and filtered. The filtered graphite was washed with 15 mL hexane. The filtrate was distilled under vacuum to remove any remaining 1,4‐dimethoxybenzene, 2‐chloro‐2‐ methylpropane and nitrobenzene (6.3 g), yielding 0.44 g 1,4‐di‐tert‐butyl‐2,5‐dimethoxybenzene (43.4% yield). What is the Atom economy, E factor, and effective mass yield?Which statement about the determination of the percentage purity of benzoic acid is correct? * A- Only sodium hydroxide solution can be used as the titrant in the volumetric analysis of benzoic acid. B- Analysing a benzoic acid sample that still contains some hydrochloric acid may yield an analyte percentage that is less than expected. C- Analysing a wet benzoic acid sample may yield an analyte percentage that is more than expected. D- All these statements are correct. E- None of these statements are correct.
- What is the corrected weight (g) of KHP (Molar mass = 204.22 g/mol) with an actual weight of 0.8026 g based on % purity of primary std. equal to 99.5% Show your complete solution.Lead nitrate Stock Solution— Dissolve 159.8 mg of lead nitrate in 100 mL of water to which has been added 1 mL of nitric acid, then dilute with water to 1000 mL. Standard Lead Solution— On the day of use, dilute 10.0 mL of Lead Nitrate Stock Solution with water to 100.0 mL. Each mL of Standard Lead Solution contains how many µg of lead? Standard Preparation— Into a 50-mL color-comparison tube pipet 2 mL of Standard Lead Solution, and dilute with water to 40 mL, and mix. Test Preparation— Into a 50-mL color-comparison tube dissolve 1.0g ascorbic acid (mol wt 176.12) in 25 mL of water, dilute with water to 40 mL, and mix. Procedure— To each of the two tubes containing the Standard Preparation, the Test Preparation, add 2 mL of pH 3.5 Acetate Buffer, then add 1.2 mL of thioacetamide–glycerin base TS, dilute with water to 50 mL. Mix, allow to stand for 2 minutes, and view downward over a white surface. The color of the solution from the Test Preparation is not darker than that of the…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+