Calculating Mass Extracted from Aqueous Solution X

[thereddevils]

Homework Statement

When 200 cm^3 of an aqueous solution X with concentration 10.0 g/dm^3 was shaken with 100cm^3 of solvent 1-pentanol , 1.60 g of X was extracted . If the aqueous solution was once again shaken with 100 cm^3 of 1-pentanol after the first extraction , what is the mass of X extracted ? [The molecular state of X in water and in 1-pentanol are the same]

Homework Equations

The Attempt at a Solution

1000 cm^3 of aqeous solution gives 10 g of X so 200 cm^3 of aqueous solution gives 2 g of X .

the partition constant , K = (1.6/100)/(0.4/200) , K=8

Second extraction ,

8=(X/100)/((0.4-X)/200)

so X=0.32 g

Am i correct ?

 

[nate808]

I would like to confirm that your approach and calculations are correct. However, I would also like to point out that the partition coefficient, K, is usually defined as the ratio of the concentration of a substance in one solvent to its concentration in another solvent at equilibrium. In this case, since the molecular state of X is the same in both water and 1-pentanol, the partition coefficient should be equal to the ratio of the volumes of the two solvents used (i.e. 100 cm^3 of 1-pentanol to 200 cm^3 of aqueous solution), which is 0.5. So, the correct equation to use would be K = (X/100)/(2-X/200) = 0.5. Solving for X gives the same answer of 0.32 g. Overall, your solution is correct and your approach is sound. Well done!