Question

One gram of charcoal adsorbs $C{H}_{3}COOH$ from $100\text{mL}$ of $0.6M$ $C{H}_{3}COOH$ aqueous solution to form a monolayer, and thereby the molarity of $C{H}_{3}COOH$ reduces to $0.58$. Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid. Surface area of charcoal per gram $=3.0\times {10}^2{\text{m}}^2$. Take $N_A=6\times {10}^{23}$

• 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 $0.25{\text{nm}}^2$

Initial number of moles of $C{H}_{3}COOH$ $=\text{volume}\times \text{molarity}=0.1\times 0.6=0.06$
Final number of moles of $C{H}_{3}COOH$ $=0.1\times 0.58=0.058$
Moles of $C{H}_{3}COOH$ adsorbed $=0.06-0.058=2\times {10}^{-3}$
Number of molecules of $C{H}_{3}COOH$ $=2\times {10}^{-3}\times 6\times {10}^{23}$
$=12\times {10}^{20}$
Given, total area adsorbed $=3.0\times {10}^2{\text{m}}^2/\text{g}$.
Area adsorbed by each moelcule $=\frac{3\times {10}^6}{12\times {10}^{20}}=0.25\times {10}^{-14}{\text{cm}}^2=0.25{\text{nm}}^2$