Water and Wastewater Treatment and Management · Water Treatment · 2 marks · Numerical answer
A 100 mL flask contains .12 mg/L geosmin. Carbon follows Freundlich $q=2.6C^{.73}$ mg adsorbate/mg adsorbent, where C is equilibrium mg/L. Find carbon dose to reach.05 mg/L, in mg/L, to three decimals.
At target concentration,$q=2.6(.05)^{.73}=.29189$ mg/mg approximately. Removed geosmin per litre is .12−.05=.07 mg/L, so carbon dose is $M/V=.07/[2.6(.05)^{.73}]=.2398151$ mg/L. Rounded **.240 mg/L**. Actual carbon mass in the.1 L flask is .0239815 mg; the requested dose is per litre.
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