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One gram of charcoal adsorbs 400 mL of ...

One gram of charcoal adsorbs 400 mL of 0.5 M acetic acid to form a mono layer and the molarity of acetic acid reduces 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)`.

Text Solution

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No of m mole of `CH_(3)COOH=100xx05=50`
Since concentriation reduces to `0.49`M
`:.` Final no. of m mole of `CH_(3)COOH=100xx0.49=49`
`:.` no of m molecules of `CH_(3)COOH` get adsorbed `=50-49=1`
`:.` No of molecules of `CH_(3)COOH` get adsorbed
`=6.02xx10^(23)xx10^(-3)=6.02xx10^(20)`
since 1g characoal has area `=3.01xx10^(2)m^(2)`
`:. 6.02 xx10^(20)` molecules of acetic acid gets adsorned in` 3.01xx10^(2) m^(2)` area
`:.` 1 molecules of acetic acid gets adsorbe `=(3.01xx10^(2))/(6.02xx10^(20))`
`=(1)/(2)xx10^(-18)=5xx10^(-19)m^(2)`
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