SATHEE: Some Basic Concepts of Chemistry

Some Basic Concepts of Chemistry - Result Question 44

####44. In a solution of $100 m L 0.5 M$ 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 $3.01 \times 10^{2} m^{2}$ , calculate the area occupied by single acetic acid molecule on surface of charcoal.

(2003)

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Solution:

Initial millimol of $C H_{3} C O O H = 100 \times 0.5 = 50$

millimol of $C H_{3} C O O H$ remaining after adsorption

$= 100 \times 0.49 = 49$

$\Rightarrow$ millimol of $C H_{3} C O O H$ adsorbed $= 50 - 49 = 1$

$\Rightarrow$ number of molecules of $C H_{3} C O O H$ adsorbed

$= \frac{1}{1000} \times 6.023 \times 10^{23} = 6.023 \times 10^{20}$

$\Rightarrow$ Area covered up by one molecule $= \frac{3.01 \times 10^{2}}{6.02 \times 10^{20}}$

$= 5 \times 10^{- 19} m^{2}$

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Some Basic Concepts of Chemistry - Result Question 44

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