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{1}{g} charcoal is placed in {100}{m}{L} of {0.5}{M}{C}{H}_{{{

Question

$1 g$ charcoal is placed in $100 m L$ of $0.5 M C H_{3} C O O H$ to form an adsorbed mono-layer of acetic acid molecule and thereby the molarity of $C H_{3} C O O H$ reduces to $0.49$ . Calculate the surface area of charcoal adsorbed by each molecule of acetic acid. Surface are of charocal $= 3.01 \times 10^{2} m^{2} / g$ .

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1 g

charcoal is placed in

100 m L

0.5 M C H_{3} C O O H

to form an adsorbed mono-layer of acetic acid molecule and thereby the molarity of

C H_{3} C O O H

reduces to

0.49

. Calculate the surface area of charcoal adsorbed by each molecule of acetic acid. Surface are of charocal

= 3.01 \times 10^{2} m^{2} / g

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Text solutionVerified

Answer:

(55.56 m o l L^{- 1})

Solution: Initial millmol of

C H_{3} C O O H = 100 \times 0.5 = 50

millmol 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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Question Text	$1 g$ charcoal is placed in $100 m L$ of $0.5 M C H_{3} C O O H$ to form an adsorbed mono-layer of acetic acid molecule and thereby the molarity of $C H_{3} C O O H$ reduces to $0.49$ . Calculate the surface area of charcoal adsorbed by each molecule of acetic acid. Surface are of charocal $= 3.01 \times 10^{2} m^{2} / g$ .
Answer Type	Text solution:1
Upvotes	150