A) Standard solution and Beer's law calibration curve1-Fill in the following table :Concentration For S.1 M. x V. =M, x V 0.000725 x 1.0ml = M, x 10.0mlSolutionAbsorbanceConcentrationS. 10.1540.5240.7890,9890.2590.0000 7250.000 2530.000 4870-0000 725???? 0.0000552S.2S.3S4ASA of unknown2- Plot a graph of Absorbance vs. conc mol/LabsorbanceAbsorbance vs. Concentration1.6141.210.80.6OY-Values0.40.22.10conc 10 mol/ LB) Analysis of unknown aspirin tablet1. Mass of aspirin tablet-2. Concentration of ASA unknown from calibration curve ..mol / L3.. Conc of Stock ASA unknown-50 mL/ 1.5 mL (D.F) x Conc of ASA unknownmol/L4. Moles of ASA in 100 mlL solution Conc of Stock ASA unknown x 0.1 L-moles5.Mass of ASA - moles ASA x Molar Mass ASA (180 g/ mol)-6.% of ASA in aspirinMass of ASA/ Mass of aspirin tablet x100 > Data Set Exp.5Use the following data to determine the masspercent of the active ingredient in Aspirina) Preparation of solutions:volume of salicylic acidvolume ofFECL3 solution1ml3.5mlSA19mlSA26.5mlSA36.5ml3.5mlSA41ml9mlThe concentration of salicylic acid= 0.000725Mb) Preparation of “STOCK ASAUNKNOWN"The mass of aspirin= 0.135The volume of stock solution of aspirin= 100mlgThe dilution factor for the stock solution= 50/2Molar mass of SAS = 180.157 g/molec) Measurements of Absorbance:SolutionAbsorbanceSA10.154SA20.524SA30.789SA40.989The unknown sample0.259Answer the following questions:1. Name the active ingredient present in theaspirin tablet.2. What is the chemical name of Aspirin tablet?3. Name the solution which is used for makingintense violet colour with ASA.

Answered: Image /qna-images/answer/db5fd2df-80f3-472e-a632-58ff4e736160.jpg

A) Standard solution and Beer's law calibration curve 1-Fill in the following table : Concentration For S.1 M¸ x V, M, x V 0.000725 x 1.0ml M, x 10.0ml Solution Absorbance Concentration S. 1 0.0000 725 S.2 0.524 0.789 0,989 0.259 0.000 253 0.000 487 0.0000725 ??? 0.0000552 S 3 S.4 ASA of unknown 2- Plot a graph of Absorbance vs. conc mol/L absorbance Absorbance vs. Concentration 16 14 1.2 0.8 0.6 OY-Values 0.4 0.2 2. 10 conc 10 mol/L B) Analysis of unknown aspirin tablet 1. Mass of aspirin tablet-

A) Standard solution and Beer's law calibration curve 1-Fill in the following table : Concentration For S.1 M¸ x V, M, x V 0.000725 x 1.0ml M, x 10.0ml Solution Absorbance Concentration S. 1 0.0000 725 S.2 0.524 0.789 0,989 0.259 0.000 253 0.000 487 0.0000725 ??? 0.0000552 S 3 S.4 ASA of unknown 2- Plot a graph of Absorbance vs. conc mol/L absorbance Absorbance vs. Concentration 16 14 1.2 0.8 0.6 OY-Values 0.4 0.2 2. 10 conc 10 mol/L B) Analysis of unknown aspirin tablet 1. Mass of aspirin tablet-

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter7: Statistical Data Treatment And Evaluation

Section: Chapter Questions

Problem 7.26QAP

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Solid Na2S is slowly added to a solution of 1.0 x 10-3 M Cr3+ (with no change in volume). At what [S2-] will Cr2S3 start to ppt? For Cr2S3, Ksp = 1.0 x 10-20
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A solution is prepared by dissolving 40.00 g of NaCl (f.w. = 58.44 g mol–1), a non-volatile solute, in enough water (m.w. = 18.02 g mol–1) to result in exactly 1 L of solution at 25 °C. Assume the density of the solution is that of pure water (dsolution = 1.000 g mL–1). The ebullioscopic constant (Kb) for water is 0.513 °C m–1. The cryoscopic constant (Kf) for water is 1.86 °C m–1. The vapor pressure of pure water is 0.0313 atm. Find the freezing point of the solution(in C to 2 decimal places)
A solution is prepared by dissolving 40.00 g of NaCl (f.w. = 58.44 g mol–1), a non-volatile solute, in enough water (m.w. = 18.02 g mol–1) to result in exactly 1 L of solution at 25 °C. Assume the density of the solution is that of pure water (dsolution = 1.000 g mL–1). The ebullioscopic constant (Kb) for water is 0.513 °C m–1. The cryoscopic constant (Kf) for water is 1.86 °C m–1. The vapor pressure of pure water is 0.0313 atm. Find the vapor pressure of the solution to 3 decimal places in atm.
A solution is prepared by dissolving 40.00 g of NaCl (f.w. = 58.44 g mol–1), a non-volatile solute, in enough water (m.w. = 18.02 g mol–1) to result in exactly 1 L of solution at 25 °C. Assume the density of the solution is that of pure water (dsolution = 1.000 g mL–1). The ebullioscopic constant (Kb) for water is 0.513 °C m–1. The cryoscopic constant (Kf) for water is 1.86 °C m–1. The vapor pressure of pure water is 0.0313 atm. Determine the following: Boiling point of solution (in °C to two decimal places) Freezing point of solution (in °C to two decimal places) Vapor pressure of the solution (in atm to three decimal places) Osmotic pressure (in atm to three decimal places)
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