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Question

One gram of charcoal adsorbs 100 mL of 0.5 M CH3COOH to form a mono-layer and thereby the molarity of acetic acid is reduced to 0.49 M. Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid. Surface area of charcoal =3.01×102 m2/gm.

A
5×1019 m2
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B
5×10+19 m2
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C
5×109 m2
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D
5×10+9 m2
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Solution

The correct option is A 5×1019 m2
Number of moles of acetic acid initially present =MV1000=0.5×1001000=0.05
Number of moles of acetic acid left =MV1000=0.49×1001000=0.049
Number of moles of acetic acid adsorbed =0.050.049=0.001 mol
Number of molecules of acid adsorbed=0.001×6.023×1023=6.023×1020
Area occupied by single molecule of acetic acid=Total areaNumber of molecules adsorbed
=3.01×1026.023×1020
=5×1019 m2



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