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Question

One gram of charcoal adsorbs CH3COOH from 100 mL of 0.6 M CH3COOH aqueous solution to form a monolayer, and thereby the molarity of CH3COOH reduces to 0.58. Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid. Surface area of charcoal per gram =3.0×102 m2. Take NA=6×1023

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 nm2
Initial number of moles of CH3COOH =volume×molarity=0.1×0.6=0.06
Final number of moles of CH3COOH =0.1×0.58=0.058
Moles of CH3COOH adsorbed =0.060.058=2×103
Number of molecules of CH3COOH =2×103×6×1023
=12×1020
Given, total area adsorbed =3.0×102 m2/g.
Area adsorbed by each moelcule =3×10612×1020=0.25×1014 cm2=0.25 nm2

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