Question

A 72.0 g sample of an organic solid is dissolved in 180mL of water. The solid is extracted using one 60 mL extraction in the first extraction of an organic solvent which has a partition (distribution) coefficient with water of 10. The first extraction removed 55.4 g of solid from water. What are the numbers that need to go in box A and B to calculate the volume of solvent (y) that would be necessary to remove an additional 7.0g from the remaining sample dissolved in water. You DON'T have to complete the calculation to solve for y.

Answer 1

Answer:

$V_{7.0}\approx 235ml$

Explanation:

From the question we are told that

mass of sample $M=72.0 grams$

volume of water $V=180 mL$

volume for extraction $V'=60mL$

partition (distribution) coefficient water $d=10$

initial extraction removal $x=55.4g$

Generally the equation for the weight of sample $x_n$  is mathematically given by

$x_n=x*(\frac{DV}{DV+V'})^n$

$x_n=55.4(\frac{10*180}{10*180+60})^1$

$x_n=53.613g$

Generally the weight extracted $x_e$ is therefore

$w_e=x-x_n$

$w_e=55.4-53613$

$w_e=1.787$

$w_e=1.787$ is extracted with 60ml solvent .

Therefore volume of solvent (y) that would be necessary to remove an additional 7.0g

$V_{7.0}=\frac{60}{1.767}*7$

$V_{7.0}=235.030ml$

$V_{7.0}\approx 235ml$