Question

$1 \,gm$ of charcoal adsorbs $100 \,mL \,0.7 \,M \,CH_3COOH$ to form a monolayer, and thereby the molarity of $CH_3COOH$ reduces to $0.59$. Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid in terms of $10^{-19}$.
Surface area of charcoal $= 3.01 \times 10^2\, m^2/gm$.

Answer 1

Number of moles of acetic acid in $100 \,mL$ before adding charcoal $= 0.07$
Number of moles of acetic acid in $100 \,ml$ after adding charcoal $= 0.059$
Number of moles of acetic acid adsorbed on the surface of charcoal.
Number of molecules of acetic acid adsorbed on the surface of charcoal
$= 0.011 \times 6.02 \times 10^{23} = 6.62 \times 10^{21}$
Surface area of charcoal $= 3.01 \times 10^2\,m^2$ (given)
Area occupied by single acetic acid molecule on the
surface of charcoal $ = \frac{3.01 \times 10^2}{6.62 \times 10^{21}} $
$= 0.45 \times 10^{-19} \,m^2$