[leena]

The organic compound Y is more soluble in diethyl ether than in water.The distribution coefficient of Y between diethyl ether and water is 4.
160ml of an aqueous solution of Y contains 7.2 g of Y.This initial aqueous solution is extracted with 80ml of diethyl ether.
The second aqueous solution so formed is seperated and then it is extracted with another 80ml of diethyl ether.
Calculate the mass of Y present in the second diethyl ether extract


[Y]ether/[Y]water = 4
Taking the mass of Y remainig in ether after the 1st extraction,as X,

[(X*1000/80)] / [(7.2-x)*1000/160] = 4
2X/(7.2-X) = 4
therfore X= 4.8g

now the mass of Y remaining in water=7.2-4.8=2.4g

Taking the mass of Y remainig in ether after the 2nd extraction,as Z,
2 Z/(2.4-Z) = 4
z= 1.6g which is the mass of Y remaining in the 2nd extract.
Is this correct?I haven't got the answers to check

[leena]

Can someone please check this for me?
I have alot more questions based on solvent extraction and I'm afraid to continue without actually knowing if what I've done is correct.

THANK YOU

[gloinddark]

i have worked out the question and i got the same answers and much the same working.

[leena]

Thank you ever so much!