InterviewSolution
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InterviewSolution
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`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` |
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Answer» `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)` `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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