CHM142 experiment5b - Mole AA = mole NaOH x RR =( )...

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OBSERVATIONS After adding crystal violet, the  resulting mixture turns to color violet  in a uniform phase By the addition of NaCl to one  of the micro test tubes, the color of  the water component turns to light  violet, leaving the other component  at its original color (violet, darker) The mixture is immiscible The water is denser than n- amyl alcohol
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II. Determination of a Distribution  Coefficient and Percentage yield Volume of NaOH  solution for  titration Original 1%  sample 39 mL Solution  extracted once 28.2 mL Solution  extracted twice 26 mL Weight of the  adipic acid in  grams Original solution 0.14235 g Solution  extracted once 0.10293 g Solution  extracted twice 0.0949 g Distribution  Coefficient Single extraction 0.383 Double extraction 0.5 Percentage Yield Percentage Yield Single Extraction 72.31 % Double extraction 66.67 % III. Computation WEIGHT OF ADIPIC ACID IN GRAMS
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Unformatted text preview: Mole AA = mole NaOH x RR =( ) massMWAA M x V NaOH x RR 1. First Titration =( . mass146 gmolAA 0 05 mnmL x 39 ) mL NaOH x 1 mn AA2 mn NaOH1 mol1000 mn Mass A = 0.14235 g 2. Second Titration =( . mass146 gmolAA 0 05 mnmL x . ) 28 2 mL NaOH x 1 mn AA2 mn NaOH1 mol1000 mn Mass B = 0.10293 g 3. Third Titration =( . mass146 gmolAA 0 05 mnmL x 26 ) mL NaOH x 1 mn AA2 mn NaOH1 mol1000 mn Mass C = 0.0949 g DISTRIBUTION COEFFICIENT 1. Single Extraction = KD A-= . B10B10 0 14235 g-. 0 10293 . = . g100 10293 g10 0 383 2. Double Extraction = KD A-= . C10C10 0 14235 g-. 0 0949 . = . g50 0949 g5 0 5 PERCENT YIELD 1. Single Extraction % = % Yield BAx100 = . . %= . % 0 10293 g0 14235 gx100 72 31 2. Double Extraction % = % Yield CAx100 = . . %= . % 0 0949 g0 14235 gx100 66 67 = KD concentration of solute in , organic solvent C1concentration of solute in , water C2...
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CHM142 experiment5b - Mole AA = mole NaOH x RR =( )...

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