Some Basic Concepts of Chemistry - Result Question 40
44. In a solution of 100 mL 0.5 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 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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Answer:
Correct Answer: 44. $\left(5 \times 10^{-19} \mathrm{~m}^2\right)$
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 $=\frac{3.01 \times 10^2}{6.02 \times 10^{20}}$
$=5 \times 10^{-19} \mathrm{~m}^2$
BETA
