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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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Number of moles of acetic acid initially present
`(MV)/(100)=(0.5xx100)/(1000) =0.05`
Number of moles of acetic acid left
`=(MV)/(1000)=(0.5xx100)/(1000)=0.049`
Number of moles of acetic acid adsorbed
=0.05-0.049=0.001 mol
Number of molecules of acid adsorbed
`=0.001xx6.023xx10^(23)=6.023xx10^(20)`
Area occupied by single molecule of acetic acid
`=("Total area")/("Number of molecules adsorbed")`
`=(3.01xx10^(2))/(6.023xx10^(20))`
`=5xx10^(-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) .

1 g charcoal is placed in 100 mL of 0.5 M CH_(3)COOH to form an adsorbed mono-layer of acetic acid molecule and thereby the molarity of CH_(3)COOH reduces to 0.49 . Calculate the surface area of charcoal adsorbed by each molecule of acetic acid. Surface are of charocal =3.01xx10^(2)m^(2)//g .

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