91Ó°ÊÓ

(a)

$\mathrm{K}=\frac{{\left[\mathrm{S}\right]}_{\mathrm{org}}}{{\left[\mathrm{S}\right]}_{\mathrm{aq}}}=\frac{{\left[\mathrm{S}\right]}_{\mathrm{oh}\mathrm{loro}\mathrm{form}}}{{\left[\mathrm{S}\right]}_{\mathrm{water}}}$

(b)

• If the volume of water is 80mL and the volume of chloroform is 10mL, we will find the quotient $(\mathrm{mol}\mathrm{S}\mathrm{in}\mathrm{chloroform})/(\mathrm{mol}\mathrm{S}\mathrm{in}\mathrm{water}):$
• Let us note that we are asked to find moles here and the equation for number of mol$\mathrm{n}=\mathrm{c}×\mathrm{V}$ , hence
  $$\begin{gathered} \frac{\left(\mathrm{molSincholoroform}\right)}{\left(\mathrm{molSinw}ater\right)}=\frac{{\mathrm{C}}_{\mathrm{chloro}\mathrm{form}}×{\mathrm{V}}_{\mathrm{chloro}\mathrm{form}}}{{\mathrm{C}}_{\mathrm{water}}×{\mathrm{V}}_{\mathrm{water}}} \\ \frac{{\mathrm{C}}_{\mathrm{chloro}\mathrm{form}}×{\mathrm{V}}_{\mathrm{chloro}\mathrm{form}}}{{\mathrm{C}}_{\mathrm{water}}×{\mathrm{V}}_{\mathrm{water}}}=\frac{0.08\mathrm{M}×10\mathrm{mL}}{0.02\mathrm{M}×80\mathrm{mL}}=0.5 \end{gathered}$$

Result:

• a)The concentration of in chloroform if ${\mathbf{\left[}\mathbf{S}\mathbf{\right]}}_{\mathbf{aq}\mathbf{}}\mathrm{is}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{020}\mathbf{M}\mathbf{}\mathrm{is}$

$$\begin{gathered} {\left[\mathrm{S}\right]}_{\mathrm{chloro}\mathrm{form}}=\mathrm{K}×{\left[\mathrm{S}\right]}_{\mathrm{water}} \\ {\left[\mathrm{S}\right]}_{\mathrm{chloro}\mathrm{form}}=4.0×0.02\mathrm{M} \\ {\left[\mathrm{S}\right]}_{\mathrm{chloro}\mathrm{form}}=0.08\mathrm{M} \end{gathered}$$

• b)the quotient $\left(\mathrm{molSinchloroform}\right)\left(\mathrm{molSinwater}\right)$is $$\begin{gathered} \frac{\left(\mathrm{molSincholoroform}\right)}{\left(\mathrm{molSinw}ater\right)}=\frac{{\mathrm{C}}_{\mathrm{chloro}\mathrm{form}}×{\mathrm{V}}_{\mathrm{chloro}\mathrm{form}}}{{\mathrm{C}}_{\mathrm{water}}×{\mathrm{V}}_{\mathrm{water}}} \\ \frac{{\mathrm{C}}_{\mathrm{chloro}\mathrm{form}}×{\mathrm{V}}_{\mathrm{chloro}\mathrm{form}}}{{\mathrm{C}}_{\mathrm{water}}×{\mathrm{V}}_{\mathrm{water}}}=\frac{0.08\mathrm{M}×10\mathrm{mL}}{0.02\mathrm{M}×80\mathrm{mL}}=0.5 \end{gathered}$$