Question

One gram of charcoal adsorbs $100mL$ of $0.5MC{H}_{3}COOH$ to form a mono-layer and thereby the molarity of acetic acid is reduced to $0.49M$. The surface area (in $m^2$) of the charcoal adsorbed by each molecule of acetic acid is $x\times {10}^{-19}{m}^2$. Surface area of charcoal= $3.0115\times {10}^2{m}^2/g$. Find the value of $x$
(Given $N_A=6.023\times {10}^{23}$)

Answer 1

$\text{Number of moles of acetic acid initially present}=\frac{MV}{1000}=\frac{0.5\times 100}{1000}=0.05$
$\text{Number of moles of acetic acid left}=\frac{MV}{1000}=\frac{0.49\times 100}{1000}=0.049$
$\text{Number of moles of acetic acid adsorbed}=0.05-0.049=0.001mol$
$\text{Number of molecules of acid adsorbed}=0.001\times 6.023\times {10}^{23}=6.023\times {10}^{20}$
$\text{Area occupied by single molecule of acetic acid}=\frac{\text{Total area}}{\text{Number of molecules adsorbed}}$
$=\frac{3.01\times {10}^2}{6.023\times {10}^{20}}$
$=5\times {10}^{-19}{m}^2$