Question

1 g of charcoal adsorbs $C{H}_{3}COOH$ from $100\text{mL}$ of $0.6MC{H}_{3}COOH$ aqueous solution to form a monolayer, and the molarity of $C{H}_{3}COOH$ reduces to $0.58$ in this process. Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid.
Surface area of charcoal = $3.0\times {10}^2{m}^2/g$

• A. $5{\text{nm}}^2$
• B. $0.5{\text{nm}}^2$
• C. $0.25{\text{nm}}^2$
• D. $2.5{\text{nm}}^2$

Answer 1

The correct option is C $0.25{\text{nm}}^2$

Milimoles of $C{H}_{3}COOH$ before adsorption $=\text{molarity}\times \text{volume}=0.6\times 100=60\text{mmol}$

Milimoles of $C{H}_{3}COOH$ after adsorption $=\text{molarity}\times \text{volume}=0.58\times 100=58\text{mmol}$

$\text{mmol}$ of $C{H}_{3}COOH$ adsorbed
= $60-58=2\text{mmol}$

Number of molecules,
$=\text{moles}\times N_A=2\times {10}^{-3}\times 6\times {10}^{23}=12\times {10}^{20}$

The area adsorbed by each molecule,
$=\frac{3\times {10}^6}{12\times {10}^{20}}$
$=0.25\times {10}^{-14}{\text{cm}}^2$
$=0.25{\text{nm}}^2$