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One gram of charcoal adsorbs 100 mL of 0...

One gram of charcoal adsorbs 100 mL of 0.5 `M CH_(3)COOH` 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 acid of charcoal `=3.01xx10^(2) m^(2)//gm`

Text Solution

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No. of moles of acetic acid initially present
`=0.5/1000 xx 100 = 0.05` mol.
No. of moles of acetic acid left after adsorption
`= (0.49)/1000 xx 100 = 0.049`.
Moles of acetic acid adsorbed `= 0.05 - 0.049 = 0.001` mol.
or = `1 xx 10^(-3)` mol.
No. of molecules of acetic acid adsorbed
` = 1 xx 10^(-3) xx 6.022 xx 10^23`
`=6.022 xx 10^20` molecules
Now, 1 g of charcoal has area `= 3.01 xx 10^2 m^2`
`:. 6.022 xx 10^20` molecules of acetic acid gets adsorbed on surface area
`:.` 1 molecule of acetic acid gets adsorbed on surface area
`= (3.02 xx 10^2)/(6.02 xx 10^20)`
`=5.0 xx 10^(-19) m^2`
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One gram of charcoal adsorbs 400 " mL of " 0.5 M acetic acid to form a mono layer and the molarity of acetic acid reduced to 0.49. Calculate the surface area of charcoal adsorbed by each molecule of acetic acid. The surface area of charcoal is 3.01xx10^(2)m^(2)g^(-1) .

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