Some Basic Concepts of Chemistry 1 Question 23
44. In a solution of $100 \mathrm{~mL} 0.5 \mathrm{M}$ acetic acid, one gram of active charcoal is added, which adsorbs acetic acid. It is found that the concentration of acetic acid becomes $0.49 \mathrm{M}$. If surface area of charcoal is $3.01 \times 10^{2} \mathrm{~m}^{2}$, calculate the area occupied by single acetic acid molecule on surface of charcoal.
(2003)
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Solution:
- Initial millimol of $\mathrm{CH}_{3} \mathrm{COOH}=100 \times 0.5=50$
millimol of $\mathrm{CH}_{3} \mathrm{COOH}$ remaining after adsorption
$$ =100 \times 0.49=49 $$
$\Rightarrow$ millimol of $\mathrm{CH}_{3} \mathrm{COOH}$ adsorbed $=50-49=1$
$\Rightarrow$ number of molecules of $\mathrm{CH}_{3} \mathrm{COOH}$ adsorbed
$$ =\frac{1}{1000} \times 6.023 \times 10^{23}=6.023 \times 10^{20} $$
$\Rightarrow$ Area covered up by one molecule $=\frac{3.01 \times 10^{2}}{6.02 \times 10^{20}}$
$$ =5 \times 10^{-19} \mathrm{~m}^{2} $$
