One gram of charcoal adsorbs 100 ml of 0.5M CH₃COOH & then molarity of acetic acid reduces to  0.49 M. The no. of milli moles of acetic acid adsorbed is ______ 

Moles of acetic acid before absorbing:

$\mathrm{Moles}=0.5\times \frac{100}{1000}=0.05 \mathrm{mol}$

Moles of acetic acid after absorbing:

$\mathrm{Moles}=0.49\times \frac{100}{1000}=0.049 \mathrm{mol}$

Moles of acetic absorbed (x) is:

$$\begin{gathered} \mathrm{x}=\mathrm{initial} \mathrm{moles}-\mathrm{final} \mathrm{moles} \\ \mathrm{x}=0.05 \mathrm{mol}-0.049 \mathrm{mol}=0.001 \mathrm{mol} \\ \mathrm{millimoles}=0.001 \mathrm{mol}\times \frac{1000 \mathrm{mmol}}{1 \mathrm{mol}}=1 \mathrm{mmol} \end{gathered}$$