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

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.

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

                                    = $[1 + 2.7(\frac{10}{100})]^{-3}$

                                    = $(1.27)^{-3}$

                                    = 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.