Question

One gram of activated charcoal was added to $100mL$ of oxalic acid solution $(0.03M)$ 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

Answer: 81

Assuming that only oxalic acid molecules get adsorbed and volume of water remains unchanged.
Number of moles of oxalic acid adsorbed on charcoal
$=\left(0.03M\times \frac{100mL}{1000mL}\right)-\left(0.021M\times \frac{100mL}{1000mL}\right)$
$=\frac{0.9}{1000}moles$
$\therefore$ Weight of oxalic acid adsorbed
$=\frac{0.9}{1000}$ moles $\times 90=81mg$
$\left[\because \text{ Molar mass of oxalic acid }=90gmo{l}^{-1}\right]$
$\therefore$ Amount of oxalic acid adsorbed per $g$ of charcoal $=81mg$