100 mL of 0.3 M acetic acid is shaken with 0.8 g of wood charcoal. The final concentration of acetic acid in the solution after adsorption is 0.125 M. The mass of acetic acid adsorbed per gram of carbon is _____________ $\times$ 10^$-$4 g.

(Given : Molar mass of acetic acid = 60 g mol^$-$1)

$$ \begin{aligned} &\text { Initial moles of acetic acid }=M_{1} \times V_{1} \\\\ &=100 \times 0.3=30 \mathrm{~m} \mathrm{~mol} . \end{aligned} $$ <br/><br/> $$ \begin{aligned} &\text { moles after adsorption }=100 \times 0.125=12.5 \mathrm{~m} \mathrm{~mol} \text {. } \\\\ &\text { moles adsorbed }=30-12.5=17.5 \mathrm{~m} \mathrm{~mol} \end{aligned} $$ <br/><br/> for acetic acid; weight adsorbed.<br/><br/> $$ \begin{aligned} &\text { moles }=\frac{\text { Weight }}{\text { Molar Mass }} \\\\ &\text { weight }=17.5 \times 10^{-3} \times 60=1050 \times 10^{-3} \mathrm{~g} \\\\ &\text { for } 0.8 \mathrm{~g} \text { charcoal mass of acetic acid adsorbed } \\\\ &=1050 \times 10^{-3} \mathrm{~g} \\\\ &1 \mathrm{~g} \text { charcoal }=\frac{1050 \times 10^{-3}}{0.8}=1312.5 \times 10^{-3} \mathrm{~g} \\\\ &=13125 \times 10^{-4} \mathrm{~g} \end{aligned} $$