Answer:
[tex]V_{7.0}\approx 235ml[/tex]
Explanation:
From the question we are told that
mass of sample [tex]M=72.0 grams[/tex]
volume of water [tex]V=180 mL[/tex]
volume for extraction [tex]V'=60mL[/tex]
partition (distribution) coefficient water [tex]d=10[/tex]
initial extraction removal [tex]x=55.4g[/tex]
Generally the equation for the weight of sample [tex]x_n[/tex] is mathematically given by
[tex]x_n=x*(\frac{DV}{DV+V'})^n[/tex]
[tex]x_n=55.4(\frac{10*180}{10*180+60})^1[/tex]
[tex]x_n=53.613g[/tex]
Generally the weight extracted [tex]x_e[/tex] is therefore
[tex]w_e=x-x_n[/tex]
[tex]w_e=55.4-53613[/tex]
[tex]w_e=1.787[/tex]
[tex]w_e=1.787[/tex] is extracted with 60ml solvent .
Therefore volume of solvent (y) that would be necessary to remove an additional 7.0g
[tex]V_{7.0}=\frac{60}{1.767}*7[/tex]
[tex]V_{7.0}=235.030ml[/tex]
[tex]V_{7.0}\approx 235ml[/tex]
