Students were tasked to conduct a recrystallization experiment to test the ability of the newly discovered solvents in their laboratory to purify and separate a mixture containing compounds X and Y. The solubility of the two compounds in g/50 mL of hot and cold solvents are given by the photo. Solubility in g/50 ml solventYSolventHotColdHotCold12.300.552.650.9523.202.904.200.5534.851.653.400.851. Given the above data, if a mixture containing 9.0 g each of X and Y is to be recrystallized from 100 milliliters of the solvents, what is the best solvent to use? Support your answer by doinga schematic diagram for EACH solvent

A question based on analytical separations that is to be accomplished.

Students were tasked to conduct a recrystallization experiment to test the ability of the newly discovered solvents in their laboratory to purify and separate a mixture containing compounds X and Y. The solubility of the two compounds in g/50 mL of hot and cold solvents are given by the photo

Students were tasked to conduct a recrystallization experiment to test the ability of the newly discovered solvents in their laboratory to purify and separate a mixture containing compounds X and Y. The solubility of the two compounds in g/50 mL of hot and cold solvents are given by the photo

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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The following evidence was obtained from an experiment to determine the solubility of calcium chloride at room temperature. A sample of saturated calcium chloride solution was evaporated to dryness, and the mass of solid residue was measured.EvidenceVolume of solution (mL) = 15.0Mass of empty beaker (g) = 90.54Mass of beaker and residue (g) = 101.36The solubility of calcium chloride is g/100 mL
A solution is prepared by dissolving 40.00 g of MgCl2 (f.w. = 95.211 g mol–1), a non-volatile solute, in enough water (m.w. = 18.02 g mol–1) to result in exactly 2 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 freezing point of the solution.
A solution is prepared by dissolving 40.00 g of MgCl2 (f.w. = 95.211 g mol–1), a non-volatile solute, in enough water (m.w. = 18.02 g mol–1) to result in exactly 2 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 osmotic pressure in atm.
Calculate the solubility at 25°C of PbCO3 in pure water and in a 0.0110M PbNO32 solution. You'll find Ksp data in the ALEKS Data tab. Round both of your answers to 2 significant digits. solubility in pure water: gL solubility in 0.0110 M Pb(NO3)2solution: gL
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. Find the osmotic pressure in atm to three 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. Determine the boiling 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. 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)
An excess amount of Ba(NO3)2 was added to 2 liters of water at 25 oC until the solution is saturated. Because Ba(NO3)2 is only slightly soluble in water, the excess Ba(NO3)2 was filtered out to get a solid-free mixture. The proponents of the experiment thought about using the concept of colligative properties, specifically boiling point elevation, to determine Ksp. It was observed in their experiments that the solution boils at 100.15 deg C. Kb of water = 0.51 K-kg/mol. Barium nitrate dissociates using: Solve the ff: a) Amount of Ba(NO3)2 dissolved in grams. MW of Ba(NO3)2 = 261.3 g/molb) Solubility product, Kspc) Vapor pressure of the solution in kPa
Although other solvents are available, dichloromethane(CH₂Cl₂) is still often used to “decaffeinate” drinks because thesolubility of caffeine in CH₂Cl₂ is 8.35 times that in water.(a) A 100.0-mL sample of cola containing 10.0 mg of caffeine is extracted with 60.0 mL of CH₂Cl₂. What mass of caffeine re-mains in the aqueous phase? (b) A second identical cola sampleis extracted with two successive 30.0-mL portions of CH₂Cl₂.What mass of caffeine remains in the aqueous phase after each extraction? (c) Which approach extracts more caffeine?
In an experiment to calculate the solubility product (Ksp) of barium nitrate (Ba(NO3)2), an excess amount of Ba(NO3)2 was added to 2 liters of water at 25oC until the solution is saturated. Because Ba(NO3)2 is only slightly soluble in water, the excess Ba(NO3)2 was filtered out to get a solid-free mixture. The proponents of the experiment thought about using the concept of colligative properties, specifically boiling point elevation, to determine Ksp. It was observed in their experiments that the solution boils at 100.15 K. Barium nitrate dissociates via the process: Ba(NO3)2(s) ⇄ Ba2+(aq) + 2NO?-3(?q) Ksp = [Ba2+][NO3−]2 Calculate the following if Kb,water = 0.51 K-kg/mol: a) Amount of Ba(NO3)2 dissolved in grams. Molar mass of Ba(NO3)2 = 261.3 g/mol b) Solubility product, Ksp c) Vapor pressure of the solution in kPa