[biocness88]

So im having an orgo lab exam tomorrow and im doing the practice problems from my book and i got stuck on this one:
The distribution coefficient, k = (conc in ligroin)/(conc in water),
between ligroin and water for solute A is 7.5.
What weight of A would be removed from a solution of 10 g of A in 100 mL of water by a single extraction with 100 mL of ligroin? What weight of A would be removed by four succe-
sive extractions with 25-mL portions of ligroin?
How much ligroin would be required to remove 98.5% of A in a single extraction?

I started off by doing this
7.5 = (x/100)/(10/100)
.75 = x/100
x = 75
i dont even know if this is right so far, but i know that once i get a number, its a percentage, right? and i think i need to use it to figure out how much of the compound is left after multiplying by the percentage and taking that amount of grams and using the equation again (and so froth, 4 times)...? its hard for me to express in words lol

---------------------


ok fine, heres an update as to what i did so far:
x/10-x = 7.5
7.5(10-x) = x
75 - 7.5 = x
75 = 8.8x
x = 8.52


heres the main part tho:
(x/25ml)/(10-x/100ml) = 7.5
basically i get x = 6.52
then i subtracted the amount that was removed- 6.52- from 10 and got 3.48

[ARGOS++]

Dear Biocness88;

If you write k = 7.5 =  c_Ligroin / c_water   - so it must mean: c_Ligroin = 7.5 * c_water.
And on the other side, after shaking: c_Ligroin + c_water = 10g in “100ml” (for the first 100ml extraction).

Conclusion:
Your Equation must be wrong and x = 75 doesn’t mean anything, - Sorry.

I hope my both equation are of some help to you.

Good Luck!
                   ARGOS⁺⁺

[Rico]

Hey biocness88

Sorry but i don't have time to go into your equations and figure out whether they are right or wrong but instead, I have attached a word-document where i have shown, how I would go about the extraction calculations.
Hope it will be of some help too you.

Rico