nB = 3.5/101.19 = 0.035 mol ; nin octaniol = x mol; nin water = 0.035 - x mol
Cin water = (0.035 - x)/0.050 mol/L
Cin octanol = x/0.100 mol/L
KD = Cin octanol/Cin water
1.45 = x*0.050/((0.035 - x)*0.100)
x = 0.026 mol
nin water = 0.035 - 0.026 = 0.0090 mol
min water = 0.0090*101.19 = 0.91 g
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