Answer :
Answer:
0.94
Explanation:
Let's assume that the total sample of the soil is 1 cm³; which has a porosity of 0.4 cm³; then the original sample of the soil will be 0.6 cm³
The mass of the soil particles in 1 cm³ of soil + water = 0.75 g
The fraction of the total atrazine can be calculated by the expression;
Total atrazine =[tex]C_w(\frac{g}{mL})*V_w(mL)+C_{(soil)}(\frac{g}{soil})*M_{(soil)}(gsoil)[/tex]
Also; [tex]C_{(soil)}(\frac{g}{soil})= K_{oc}(\frac{mL}{gOC})*f_{oc} (\frac{gOC}{gsoil})*C_w(\frac{g}{mL})[/tex]
NOTE THAT: [tex]K_{oc}( \frac{mL}{gOC})=K_{oc}\frac{L}{kgOC}[/tex]
Total atrazine can now be calculated as:
[tex]C_w*0.4+K_{oc}f_{oc}C_w*0.75[/tex] = [tex]\frac{C_{soil}*M_{soil}}{Total atrazine}[/tex]
[tex]=\frac{K_{oc}f_{oc}C_w*M_{soil}}{C_w*0.4+K_{oc}f_{oc}C_w*0.75} = \frac{0.75*K_{oc}f_{oc}}{0.4+K_{oc}f_{oc}*0.75}[/tex]
[tex]= \frac{10^{2.52}*0.025*0.75}{0.4+10^{2.52}*0.025*0.75}[/tex]
= 0.94
Thus,the fraction of total atrazine that will be adsorbed to the soil = 0.94.