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Question

2.0 g of charcoal is placed in a 100 mL solution of 0.05 M CH3COOH to form an adsorbed monoacidic layer of acetic acid molecules and thereby the molarity of CH3COOH reduces to 0.49 M. The surface area of charcoal is 3×102 m2g1. The surface area of charcoal used by each molecule of acetic acid is:

A
1.0×1018 cm2
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B
1.0×1019 cm2
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C
1.0×1013 cm2
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D
1.0×1014 cm2
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Solution

The correct option is C 1.0×1014 cm2
CH3COOH absorbed = 0.5 - 0.49 = 0.01M
No of molecules absorbed
= 0.01×1001000×NA=6×1020
Total area of charcoal =2×3×102
= 600 m2
Area per molecule = 6006×1020
= 1×1018 m2
= 1×1018×104 cm2
= 1×1014 cm2

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