Starting with 2.0 g of hand sanitizer in 10 mL of methanol (density of methanol = 0.791 g/mL), the Karl Fischer titrator returned a water content of 4.0%. What was the original water concentration in the hand sanitizer?
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- An unknown sample containing mixed alkali (NAOH, NaHCO3, or NazCO3) was analyzed using the double flask method. A 250 mg sample was dissolved in 250 mL CO, free water. A 20.0 mL aliquot of this sample required 11.3 mL of 0.009125 M HCI solution to reach the phenolphthalein end point. Another 29.0 mL aliquot of the sample was titrated to the bromocresol green endpoint using 311 mL of the standard acid. What is/are the component/s of the sample? O Cannot be determined O NAHCO3 only O NAOH and Na,CO3 O NazCO3 only O NAHCO, and Na2CO3An unknown sample containing mixed alkali (NAOH, NaHCO3, or NazCO3) was analyzed using the double flask method. A 250 mg sample was dissolved in 250 mL CO, free water. A 20.0 mL aliquot of this sample required 11.3 mL of 0.009125 M HCI solution to reach the phenolphthalein end point. Another 29.0 mL aliquot of the sample was titrated to the bromocresol green endpoint using 31.1 mL of the standard acid. What is/are the component/s of the sample? O NazCO3 only NaHCO3 only O NAHCO3 and Na>CO3 NaOH and NazCO3 O Cannot be determinedAn unknown sample containing mixed alkali (NAOH, NaHCO3, or NazCO3) was analyzed using the double flask method. A 250 mg sample was dissolved in 250 mL CO2 free water. A 20.0 mL aliquot of this sample required 11.3 mL of 0.009125 M HCI solution to reach the phenolphthalein end point. Another 29.0 mL aliquot of the sample was titrated to the bromocresol green endpoint using 31.1 mL of the standard acid. How many millimoles of the components are there in the original solid sample? O 0.28 mmol Na,CO3 O 1.29 mmol NaOH, 0.97 mmol Na2CO3 O 0.103 mmol NaOH, 0.078 mmol NazCO3 O 1.29 mmol Na2CO3, 0.97 mmol NaHCO3 O Cannot be determined
- An unknown sample containing mixed alkali (naoh, nahco3, or na2co3) was analyzed using the double flask method. a 250 mg sample was dissolved in 250 ml co2 free water. a 20.0 ml aliquot of this sample required 11.3 ml of 0.009125 m hcl solution to reach the phenolphthalein end point. another 29.0 ml aliquot of the sample was titrated to the bromocresol green endpoint using 31.1 ml of the standard acid. how many millimoles of the components are there in the original solid sample10-27. A 10.231-g sample of window cleaner containing ammonia was diluted with 39.466 g of water. Then 4.373 g of solution were titrated with 14.22 mL of 0.106 3 M HCl to reach a bromocresol green end point. (a) What fraction of the 10.231-g sample of window cleaner is contained in the 4.373 g that were analyzed? (b) How many grams of NH3 (FM 17.031) were in the 4.373-g sample? (c) Find the weight percent of NH3 in the cleaner.The benzoic acid extracted from 105.6 g of catsup required a 14.67 mL titration with 0.0514 N NaOH. Express the results of this analysis in terms of percent sodium benzoate. MW=144.10 g/mol Answer: % Na benzoate = Blank 1%
- A student performed three KHP titrations in Part 1 in order to standardize their NaOH solution. They reported the following measurements: Mass of KHP + Vial Mass of 'Empty' Vial Final Burette Reading Initial Burette Reading Determination #1 Determination #2 Determination #3 14.1992 g 14.0219 g 14.0982 g 13.3149 g 13.2301 g 13.2779 g 39.21 mL 0.49 mL 35.46 mL 0.86 mL 37.42 mL 1.59 mL How many moles of NaOH did they use in their third titration? Report your answer to the correct number of significant figures and only report the numerical value (no units).2+ as (6) A sample of an ore was analyzed for Cu* follows. A 1.25 g sample of the ore was dissolved in acid and diluted to volume in a 250 mL volumetric flask. A 20 mL portion of the resulting solution was transferred by pipet to a 50 ml volumetric flask and diluted to volume. An analysis showed that the concentration of 2+ Cu* in the final solution was 4.62 ppm. What is the weight percent of Cu in the original ore?A typical protein contains 16.2 wt% nitrogen. A 0.500-mL aliquot of protein solution was digested, and the liberated NH3 was distilled into 10.00 mL of 0.021 40 M HCl. Unreacted HCl required 3.26 mL of 0.019 8 M NaOH for complete titration. Find the concentration of protein (mg protein/mL) in the original sample.
- The ethyl acetate concentration in an alcoholic solution was determined by diluting a 10.00-mL sample to 100.00 mL. A 20.00-mL portion of the diluted solution was refluxed with 40.00 mL of 0.04672 M KOH:CH3COOC2H5 + OH- → CH3COO- + C2H5OHAfter cooling, the excess OH2 was back-titrated with 3.41 mL of 0.05042 M H2SO4. Calculate theamount of ethyl acetate (88.11 g/mol) in the original sample in gramsTITRIMETRIC DATA SAMPLE: CANE VINEGAR % acidity in label: 4.5% %purity of KHP: 99.80% FORMULA WEIGHT of KHP: 204.22 g/mol STANDARDIZATION OF NaOH SOLUTION TRIAL 1 TRIAL 2 TRIAL 3 Weight of KHP, g 0.1012 0.1004 0.09987 Initial volume of NaOH, mL 5.00 10.01 Final volume of NaOH, mL 4.95 9.99 14.93 Molarity of NaOH Average Molarity of NaOH ANALYSIS OF ACETIC ACID IN A VINEGAR SAMPLE TRIAL 1 TRIAL 2 TRIAL 3 Volume of vinegar, mL Initial volume of NaOH, mL 1.00 1.00 1.00 14.98 22.90 30.87 Final volume of NaOH, mL 22.84 30.77 38.75 Molarity of acetic acid Average molarity of acetic acid ANALYSIS OF CARBONIC ACID IN A SODA SAMPLE TRIAL 1 TRIAL 2 TRIAL 3 Volume of soda, mL 20.0 20.0 20.0 Initial volume of NaOH, mL 20.20 22.05 23.93 Final volume of NaOH, mL 22.03 23.89 25.8 Molarity of carbonic acid Average molarity of carbonic acidA 0.5131-g sample that contains KBr (MM: 119.0023) is dissolved in 50 mL of distilled water. Titrating with 0.04614 M AgNO3 requires 25.13 mL to reach the Mohr end point. A blank titration requires 0.65 mL to reach the same end point. Report the %w/w KBr in the sample.