Assume that 1.00 mL (1.02 g) of crude product mixture is obtained from the reactionbefore the washing steps, and the distribution constant of the product mixture in brine isK = (Cmixture/Cbrine) = 50.0. Assuming that after mixing the product mixture volume isstill 1.00 mL (i.e., the volume lost is small), and the volume of the first brine wash layeris 2.00 mL. What mass (grams) of product mixture is lost (dissolves) in the first brinewash? (See Technique 13.2, p 53.) Show work. 5413.2 LiquidsA similar washing procedure can be applied to liquid samples. Suppose that a 1.00 Laqueous solution contains 100. g of solute A and 10.0 g of solute B (an impurity). Suppose thatsolute A has a distribution coefficient KA = 99.5, and that solute B has a distribution coefficientKB = 0.100 for the solvent ether. That is, A is much more soluble in ether than in water, while Bis more soluble in water than in ether. If the aqueous solution were extracted once with 1.00 L ofether, the ether layer would contain nearly all of the solute A (99.0 g), and a little solute B (0.909КА = 99.5 =KB = 0.100 =KA= 99.5 =CatherCwaterKB = 0.100 =CetherCaterCetherCaterCether=Cwatera1.00 L(100.- a)1.00 Lb1.00 L(10.0 - b)1.00 LThese values are obtained by solving the respective distribution coefficient equations above for aand b, respectively, where a is defined as grams of solute A dissolved in 1.00 L of ether, and b isgrams of solute B dissolved in 1.00 L of ether. The layers are then separated. (In a separatoryfunnel, the aqueous layer is drained out, and the ether layer remains in the funnel). Suppose that1.00 L of pure water (wash) is then added, and the mixture is shaken vigorously until equilibriumis achieved. At this point the ether layer contains 98.0 g of solute A and only 0.0827 g of soluteB. (These values were obtained by solving the following equations for a' and b', respectively,where a' is grams of solute A lost into the aqueous layer, and b' is grams of solute B lost into theaqueous layer.) As an exercise, you should verify the 98.0 g of solute A, and the 0.0827 g ofsolute B obtained above by working out the algebra.(99.0 - a')1.00 La'1.00 L(0.909 - b')1.00 Lb'1.00 LChem 2315 Lab Manual==a 1.00 L(100. a) 00 Lb 1.00 L(10.0-b) 00 L(99.0 - a') 00 La' 00 L(0.909 - b')00 Lb' 1.00 L==a(100.- a)b(10.0 - b)(99.0 - a')a'(0.909 - b')b'

Mass of crude product = 1.02 gThe partition coefficient for the extraction, K = 50.0Volume of…

Assume that 1.00 mL (1.02 g) of crude product mixture is obtained from the reaction before the washing steps, and the distribution constant of the product mixture in brine is K = (Cmixture/Cbrine) = 50.0. Assuming that after mixing the product mixture volume is still 1.00 mL (i.e., the volume lost is small), and the volume of the first brine wash layer is 2.00 mL. What mass (grams) of product mixture is lost (dissolves) in the first brine wash? (See Technique 13.2, p 53.) Show work.

Assume that 1.00 mL (1.02 g) of crude product mixture is obtained from the reaction before the washing steps, and the distribution constant of the product mixture in brine is K = (Cmixture/Cbrine) = 50.0. Assuming that after mixing the product mixture volume is still 1.00 mL (i.e., the volume lost is small), and the volume of the first brine wash layer is 2.00 mL. What mass (grams) of product mixture is lost (dissolves) in the first brine wash? (See Technique 13.2, p 53.) Show work.

EBK A SMALL SCALE APPROACH TO ORGANIC L

4th Edition

ISBN:9781305446021

Author:Lampman

Publisher:Lampman

Chapter79: Solubility

Section: Chapter Questions

Problem 1P

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