Question

A compound distributes between benzene (solvent 1) and water (solvent 2) with a distribution coefficient, K = 2.7. If 1.0g of the compound were dissolved in 100mL of water, what weight of compound could be extracted by THREE sequential 10-mL portions of benzene?

Answer 1

Step-by-step explanation:

It is given that solvent 1 is benzene and solvent 2 is water. Value of
`K_(p)` = 2.7.

Volume of solvent 1 (
`V_{S_(1)}`) = 100 mL.

Volume of solvent 2 (
`V_{S_(2)}`) = 10 mL.

Therefore, calculate the remaining fraction as follows.

`f_(n)` =
`[1 + K_(p)(\frac{V_{S_(2)}}{V_{S_(1)}})]^(-n)`

=
`[1 + 2.7((10)/(100))]^(-3)`

=
`(1.27)^(-3)`

= 0.488

Since, mass of compound is given as 1 gram.

So, reamining extract will be 1 -
`f_(n)` = 1 - 0.488 = 0.512.

Thus, we can conclude that the weight of compound extracted is 0.512.