2. Suppose we extract an aqueous solution which is very dense (because it contai

Question

2. Suppose we extract an aqueous  solution which is very dense (because it contains heavy salts) with methylene chloride and, when the layers separate, we are not sure which is the aqueous layer and which is the methylene chloride layer. What simple experimental procedure could be used to settle this problem?

3. (a) A 5.00-g sample of compound A was dissolved in 50 ml of water. This solution was then shaken with 50 ml of ether. At this point the ether was found to contain 2.00 g of compound A. Calculate the partition ratio:  [A]ether / [A]water= alpha

(b) The aqueous solution was then extracted four more times with 50-mL portions of ether (a total of 5 extractions). How much of the compund remained in the aqueous layer? How much of compound A was in the five combined ether layers? What percentage of A was extracted from the water?

(c) How much A would have been extracted into ether if the orginal aqueous solution had been extracted once with a 500-mL portion of ether? What percentage of A would have been extracted by the ether?

Explanation / Answer

2. uper layer is water and downlayer is methylene chloride

3. a) alpha= 2/3 = .667

b) if we extracted n times with Vo volume organic solvent and aquos layer volume is Va and partition ratio is ALPHA then fraction remaining in aquos layer q wil be

qn = ( 1/ ( ALPHA*VR+1))n where VR = Vo/ Va

VR= 50/50 = 1

n= 5 ALPHA= .667

q5= .077 ( PUTING ALL THE VALUE IN EQUATION)

SO, IN AQOUS LAYER 5*.077 =.388g WILL BE REMAIN

IN ORGANIC LAYER ( 5-.388) = 4.611 g

% EXTRACTED IS ( 1- .077) * 100 = 92.3 %

C)

USING SAME FORMULA WE WILL CALCULATE

VR= 500/50 = 10

n = 1 ALPHA= .667

q1 = .1303 ( PUTTING ALL THE VALUE IN BOLDED EQUATION IN PREVIOUS ANSWER)

% EXTRACTED IS (1- .1303)* 100 = 86.96%

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