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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 m2g−1. 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 D 1.0×10−14 cm2CH3COOH 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 = 1×10−18 m2 = 1×10−18×104 cm2 = 1×10−14 cm2

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