50 ml of 1 M acetic acid is shaken with 0.5 g wood charcoal. The final conc. of the solution after adsorption is 0.5 M. What is the amount of acetic acid adsorbed per gram of carbon? 

Molarity of acetic acid before absorption:

$$\begin{gathered} \mathrm{Molarity}=\frac{\mathrm{mass}}{\mathrm{molar} \mathrm{mass}}\times \frac{1000 }{\mathrm{V} (\mathrm{in} \mathrm{mL})} \\ 1 \mathrm{M}=\frac{\mathrm{mass}}{60}\times \frac{1000 }{50 \mathrm{mL}} \\ \mathrm{mass}=3 \mathrm{g} \end{gathered}$$

Molarity of acetic acid after absorption:

$$\begin{gathered} \mathrm{Molarity}=\frac{\mathrm{mass}}{\mathrm{molar} \mathrm{mass}}\times \frac{1000 }{\mathrm{V} (\mathrm{in} \mathrm{mL})} \\ 0.5 \mathrm{M}=\frac{\mathrm{mass}}{60}\times \frac{1000 }{50 \mathrm{mL}} \\ \mathrm{mass}=1.5 \mathrm{g} \end{gathered}$$

Mass of acetic acid absorbed:

$\mathrm{Mass} \mathrm{of} \mathrm{acetic} \mathrm{absorbed}=3.0 \mathrm{g}-1.5 \mathrm{g}=1.5 \mathrm{g}$

If 1.5 g of acetic acid absorbed in 0.5 g of charcoal, then mass of oxalic acid absorbed in 1 g of charcoal:

$$\begin{gathered} \mathrm{Mass} \mathrm{of} \mathrm{absorbed} \mathrm{charcoal}=1.5 \mathrm{g}\times 2 \\ =3.0 \mathrm{g} \end{gathered}$$