Question

Three mol of pesticide of molecular mass 200 g/mol is applied to a closed system containing 20 m3 of water, 10 m3 of air, 1 m3 of sediment, and 0.001 m3 of biota (fish). If the concentration ratios are air/water 0.1; sediment/water 50; and biota/water 200, what are the concentrations and amounts in each phase in both grams and mole units? (cw = 8.4 g/m3 )

Answer 1

Cw=8.4g/m3 => concentration of pesticide in water

Cw=8.4g/m3 => Molw= 8.4(g/m3)*20m3=(168g of pesticide in water)/200(g/mol)=0.84mol of pesticide in water

Cair/Cw=0.1 => Cair=0.1*Cw => Cair= 0.1*8.4=0.84g/m3 => Molair=0.84(g/m3)*10m3=(8.4g of pesticide in air)/200(g/mol)=0.042mol of pesticide in air

Csed/Cw=50 => Csed=50*Cw => Csed=50*8.4=420g/m3 => Molsed=420(g/m3)*1m3=(420g of pesticide in sediment)/200(g/mol)=2.1mol of pesticide in sediment

Cbio/Cw=200 => Cbio=200*Cw => Cbio=200*8.4=1680g/m3 =>Molbio=1680(g/m3)*0.001m3=(1.680g of pesticide in biota)/(200g/mol)= 0.008mol of pesticide in biota