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Question

# 1 g of charcoal adsorbs CH3COOH from 100 mL of 0.6 M CH3COOH aqueous solution to form a monolayer, and the molarity of CH3COOH reduces to 0.58 in this process. Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid. Surface area of charcoal = 3.0×102 m2/g

A
5 nm2
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B
0.5 nm2
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C
0.25 nm2
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D
2.5 nm2
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Solution

## The correct option is C 0.25 nm2Milimoles of CH3COOH before adsorption =molarity×volume=0.6×100=60 mmol Milimoles of CH3COOH after adsorption =molarity×volume=0.58×100=58 mmol mmol of CH3COOH adsorbed = 60−58=2 mmol Number of molecules, =moles×NA=2×10−3×6×1023=12×1020 The area adsorbed by each molecule, =3×10612×1020 =0.25×10−14cm2 =0.25 nm2

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