Question

The solute in Problem 23 - 8 is initially dissolved in 80.0 mL of water. It is then extracted six times with 10.0mL portions of chloroform. Find the fraction of solute remaining in the aqueous phase.

Short answer

$q^n={\left(\frac{V_1}{V_1+K{V}_2}\right)}^n$

$q^n={\left(\frac{80mL}{80mL+\left(4.0\times 10mL\right)}\right)}^6$is the fraction of solute remaining in the aqueous$q^n=0.088$ 8phase q .

Step-by-step solution

- Procedure

• In this task it is given that the solute in Problem 23 - 8 is initially dissolved in of water which is then further extracted six times with 10.0mL portions of chloroform.
• Let us find the fraction of solute remaining in the aqueous phase.

– Listing of known values

$$\begin{gathered} V_1=80mL \\ {V}_2=10mL \\ K=4.0 \\ n=6times \end{gathered}$$

Now, let us find the fraction remaining $\left(q^n\right)$

– calculation of qn 

The value of $\left(q^n\right)$ is calculated below.

$$\begin{gathered} q^n={\left(\frac{V_1}{V_1+K{V}_2}\right)}^n \\ {q}^n={\left(\frac{80mL}{80mL+\left(4.0\times 10mL\right)}\right)}^6 \\ {q}^n=0.088 \end{gathered}$$

Result:

The fraction of solute remaining in the aqueous phase is 0.88