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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)`. |
Answer» Initial millomoles of `CH_(3)COOH=100xx0.5=50` Final millimoles of `CH_(3)COOH=100xx0.49=49` millimoles adsorbed `=50-49=1` moles adsobed `=(1)/(1000)` Molecules adsorbed`=(1)/(1000)xx6.023xx10^(23)` `=6.023xx10^(20)` Area per molecule`=("Total area")/("Number of molecules")` `=(3.01xx10^(2))/(6.023xx10^(20))=5xx10^(19)m^(2)` |
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