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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 `0.05 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 `3xx10^(2) m^(2)g^(-1)`. The surface area of charcoal is adsorbed by each molecule of acetic acid is

A

`1.0xx10^(-18)m^(2)`

B

`1.0xx10^(-19)m^(2)`

C

`1.0xx10^(13)m^(2)`

D

`1.0xx10^(-22)m`

Text Solution

Verified by Experts

The correct Answer is:
A
`CH_(3)COOH` adsorbed `=0.5-0.49=0.01M`
Number of molecules adsorbed `=0.01xx(100)/(1000)xx6xx10^(23)=6xx10^(20)`
Total area of charcoal `=2xx3xx10^(2)=600m^(2)`
implies Area per molecule `=(600)/(6xx10^(20))=1xx10^(-18)m^(2)`
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