I am at work right now and i was told to post this questionagain. Please help. Question # 4:The distribution coefficient, k= (conc. In ligroin/conc. I

Isa Trevino

Isa Trevino

Answered question

2021-01-19

I am at work right now and i was told to post this questionagain. Please help. Question # 4:The distribution coefficient, k= (conc. In ligroin/conc. In water),between ligroin and water for solute A is 7.5. What weight of Awould be removed from a solution of 10 grams of A in 100 mL ofwater by single extraction with 100 mL of ligroin? What weight of Awould be removed by four successive extractions with 25 mL portionsof ligroin? How much ligroin would be required to remove 98.5 % ofA in a single extraction?

Answer & Explanation

Viktor Wiley

Viktor Wiley

Skilled2021-01-20Added 84 answers

Q1 7.5=1100(10x)100
This gives x=8.82 grams extracted into ligroin phase with onepass.
Q2:
Here you divide up the 100 mL of ligroin and use 25 mL freshphases in 4 passes on the aqueous phase
7.5=x25(10x)100=4x10x
On first pass, x is determined to be 6.52 g. Nowyou have 10-6.52 g remaining in aqueous layer or 3.48 g
For the second pass with a new 25 mL fresh ligroin phase
7.5=4x3.48x
Determine the new value of x and add this to 6.52....this sumis the amount extracted after two passes. Carry out thesame processes for the 3rd and 4th passes. The key is tocompare the total extracted in the 4 pass approach to the earlier 1pass approach to see which process effects the best separation.
Q3:
In the last question, you are told how much of A remains in theaqueous layer and how much remains in the ligroin layer. Youalso know the volume of water and you need to figure out the volumeof ligroin, Y.
7.5=9.85Y0.15100
Solve for Y. This is the volume of ligroin in mL to extract 98.5% of A in 1 pass. You will find this to be substantiallylarger than 100 mL!

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