Question

$Onegram$ of charcoal adsorbs $100mL$ of $0.5MC{H}_{3}COOH$ to form a monolayer and thereby the molarity of acetic acid is reduced to $0.49M$. Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid. Surface area of charcoal $=3.01\times {10}^2{m}^2/gm$.

Answer 1

Moles of $C{H}_{3}COOH$ absorbed $=$ Initial moles -Final moles.

$=0.5\times 100\times {10}^{-3}-0.049\times 100\times {10}^{-3}$

$=0.05-0.049=0.001$

No. of molecules of $C{H}_{3}COOH$absorbed $=0.001\times 6.02\times {10}^{23}$

$=6.02\times {10}^{20}$

Surface area $=3.01\times {10}^2{m}^2$ (given)

Surface area of charcol absorbed by in each molecule of $C{H}_{3}COOH=\frac{3.01\times {10}^2}{6.02\times {10}^{20}}=5\times {10}^{-19}{m}^2$