Question

One gram of activated charcoal was added to $100mL$ of oxalic acid solution $\left(\right.0.03M\left.\right)$ in a flask. After an hour, it was filtered and the strength of the filtrate was found to be $0.021M.$ The amount of oxalic acid adsorbed (per gram of charcoal) is_______ $mg.$

Answer 1

Assuming that only oxalic acid molecules get adsorbed and volume of water remains unchanged.
Number of moles of oxalic acid adsorbed on charcoal
$=\left(0 . 03 M \times \frac{100 mL}{1000 mL}\right)-\left(0 . 021 M \times \frac{100 mL}{1000 mL}\right)$
$=\frac{0 . 9}{1000}moles$
$\therefore $ Weight of oxalic acid adsorbed
$=\frac{0 . 9}{1000}$ moles $\times 90=81mg$
$\left[\because \text{ Molar mass of oxalic acid } = 90 gmol^{- 1}\right]$
$\therefore $ Amount of oxalic acid adsorbed per $g$ of charcoal $=81mg$