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2.0 g of charcoal is placed in 100 mL of...

2.0 g of charcoal is placed in 100 mL of 0.5 M `CH_3`COOH to form an adsorbed mono-acidic layer of acetic acid molecules and thereby the molarity of `CH_3` COOH reduces to 0.49. The surface area of charcoal is `3 xx 10^2 m^2 g^(-1)`. The surface area of charcoal adsorbed by each molecule of acetic acid is

A

`1.0 xx 10^(-18) m^(2)`

B

`1.0 xx 10^(-19) m^(2)`

C

`1.0 xx 10^(13) m^(2)`

D

` 1.0 xx 10^(22)` m

Text Solution

Verified by Experts

The correct Answer is:
A
`CH_(3)COOH` adsorbed = 0.5 - 0.49 = 0.01 M
Number of molecules adsorbed `=0.01xx100/1000xx6xx10^(23)=6xx10^(20)`
Total are of charcoal =`2xx3xx10^(2) =600 m^(2) rarr` Area per molecule =`(600)/(6xx10^(20))=1xx10^(-18)m^(2)`
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