Question

One gram of charcoal absorbs 100 ml 0.5 M $C{H}_{3}COOH$ to form a mono layer, and thereby the molarity of $C{H}_{3}COOH$ reduces to 0.49. Calculate the surface area of the charcoal absorbed by each molecule of acetic acid. Surface area of charcoal = $3.01\times {10}^2{m}^2/gm$

Answer 1

Number of moles of acetic acid = $concentration\times volume$

= $0.5mol{L}^{-1}$$\times 0.1L$

=$0.05moles$

Number of moles of acetic acid left after adsorption$=0.49mol$ $L^{-1}$ $\times 0.1L$

$=0.049moles$

Number of moles of acetic acid adsorb on surface$=0.05-0.049=0.001mol$

Number of moles of molecules adsorb on surface$=0.001\times 6.022\times {10}^{23}$

=$\times 6.022\times {10}^{20}$

Surface Area occupied by each molecule = total surface area of charcoal / number of molecules of acetic acid

= $3.01\times {10}^2{m}^2{g}^{-1}$ / $6.022\times {10}^{20}$

$5.0\times {10}^{-19}{m}^2{g}^{-1}$