Q3/ A) 0.63 g of a sample containing Na CO,, NaHCO, and inert impurities is titrated with 0.2M HCI, requiring 17.2 ml to reach the phenolphthalein end point and a total of 43.5 mL. to reach the modified methyl orange end point, How many grams Na CO, and NAHCO, are in the mixture?
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- The concentration of Cl– in a 300.0-mL sample of water drawn from a fresh water acquifer suffering from encroachment of sea water, was determined by titrating with 0.0516 M Hg(NO3)2. The sample was acidified and titrated to the diphenylcarbazone end point, requiring 6.18 mL of the titrant. Report the concentration of Cl–in parts per millionWhat is the percentage purity of acetic acid if 2.6 grams required 32.5 ml of 0.994 N NaOH solution to reach the endpoint? Does it conform to USP requirements for acetic acid 3%-6%?A 1.5000-g sample of cereals was analyzed for nitrogen using the Kjeldahl procedure. The receiving flask contained 69 mL of 0.02 M HCl. After the ammonia was collected, the solution was titrated with 0 M NaOH, requiring 11.72 mL to reach the methyl red endpoint. Calculate and percent protein in the sample (f=5.70). Ans. in 3 SFs
- 2.) A 0.2185g sample of NaCl was assayed using Volhard Method with 50mL 0f 0.0998N AgNO3 and 11.9mL of 0.1350N NH4SCN. Calculate the % NaCl in the sample.20 aspirin tablets labeled 80mg were dissolved in 100mL of 90% ethanol. A 10mL aliquot was taken and was used for assay. The analyte followed usual process and was treated with 50mL of 0.1000N NaH and was titrated with 35mL 0.1050N H2O4 until the solution achieved completion. Calculate the % content of the total aspirin capsules and the actual label claimA sample is analyzed for chloride by the Volhard method. From the following data, calculate the percentage of chloride present:Weight of sample = 6.0000 g dissolved and diluted to 200 mLAliquot used = 25.00 mL AgNO3 added = 40.00ml of 0.1234MKSCN for back titration = 13.20ml of 0.0930M
- A 0.1220 g vitamin C tablet was dissolves in acid. A 25.00 ml aliquot of 0.01740 M KIO3 was added along with excess KI. The resulting solution was titrated with 0.07210 M Na2S2O3, requiring 21.44 mL to reach the endpoint. calculate percent by weight of ascorbic acid in the tablet.1. A 1.2-gram sample of lanolin was treated with Wij’s solution and excess potassium iodide solution. The liberated iodine reacted with 30 ml of 0.1 N sodium thiosulfate solution. If the iodine value was determined as 12.69, what is the volume used in blank titration? 2. A fat sample with combination of acids contain standard hydrochloric acid for blank and sample with 8mL and 5mL respectively. The normality of the standard hydrochloric acid is 0.93N and the weight of the sample is 3 grams. Calculate the saponification value. 3. A 3.50-gram sample of Streptomycin powder was tested for its water content. If the water equivalence factor of the KF reagent was 4.6, what is the percentage water content of the sample if 9.2 ml of the KF reagent was used? 4. A 500mg oil sample is taken from a conical flask and is dissolved in 50mL distilled alcohol. An indicator is added and is then titrated against 0.112N KOH until a slight pink color appears. It took 17.6mL of the titrant to reach the…Zhongli is a speleologist tasked to analyze the CaCO₃ content of a limestone stalactite. A 5.0000-g sample was dissolved in 25.00 mL of 1.350 M HCl, it was then heated to expel any CO₂ formed. The excess HCl was titrated to a phenolphthalein end point, it used 37.50 mL of 0.1200 M NaOH. A. How many moles of HCl was added initially to digest the limestone sample? B. How many moles of CaCO₃ is present in the limestone sample? C. What is the purity of the limestone in terms of %w/w CaCO₃?
- I'm trying to find the wt% of an unknown sample of KHP. I titrated a 1.4024 g of the unknown KHP dissolved in 50mL of water with 0.1327 M NaOH and the endpoint was at 13.02 mL of NaOH.A 10.00mL sample of alcoholic ethyl acetate was diluted to 100.00 mL. 20.00 mL was aliquoted and mixed with 40.00 mL of 0.04672 M KOH. The resulting mixture was heated for 2 hours. CH3COOC2H5 + OH- → CH3COO- + C2H5OH After cooling, the excess OH was back titrated with 3.41 mL of 0.05042 M H2SO4. Answer the following: Calculate the number of moles of OH- that reacted with ethyl acetate. Calculate the number of moles of ethyl acetate in the 20.00 mL solution. What is the mass of ethyl acetate (FW=88.11 g/mol) in the original 10.00 mL sample?The molar solubility of MgCO3 (Ksp = 3.50 x 10-8) in distilled water at room temperature is ______ M. Titration of a 50.00-mL aliquot of the saturated solution will require ______mL of 0.005000 M HCl to reach the phenolphthalein endpoint.