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

One gram of charcoal adsorbs `100ml` of `0.5M` acetic acid to form a monolayer, 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

Verified by Experts

The correct Answer is:
`5 xx 10^(-9) m^(2)`
Final molarity `= 5 - 49 = 01 M`
mole `= M xx v = 0.1 xx (100)/(1000)`
`= 10^(-3)`
no of molecule = moles `xx N_(A)`
`= 10^(-3) xx N_(A) = 6.02 xx 10^(20)`
1gm contain charcoal `= 3.01 xx 10^(2) m^(2)`
`6.02 xx 10^(20)` molecule of acetic acid absorbed charcoal `= 3.01 xx 10^(2)`
1 molecule of acetic acid adsorbed charcoal `= 3.01 xx 10^(2) m^(2)`
`6.01 xx 10^(20)` molecule of acetic acid absorbed charcoal `= 3.01 xx 10^(2)`
molecule of acetic acid adsorbed charcoal `= (3.01 xx 10^(2))/(6.02 xx 10^(20)) = 5 xx 10^(-19)`
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