Question

$2.0\text{g}$ of charcoal is placed in a $100\text{mL}$ solution of $0.05\text{M}C{H}_{3}COOH$ to form an adsorbed monoacidic layer of acetic acid molecules and thereby the molarity of $C{H}_{3}COOH$ reduces to $0.49\text{M}$. The surface area of charcoal is $3\times {10}^2{\text{m}}^2{\text{g}}^{-1}$. The surface area of charcoal used by each molecule of acetic acid is:

• A. $1.0\times {10}^{-18}{\text{cm}}^2$
• B. $1.0\times {10}^{-19}{\text{cm}}^2$
• C. $1.0\times {10}^{-13}{\text{cm}}^2$
• D. $1.0\times {10}^{-14}{\text{cm}}^2$

Answer 1

The correct option is D $1.0\times {10}^{-14}{\text{cm}}^2$

$C{H}_{3}COOH$ absorbed = 0.5 - 0.49 = 0.01M
No of molecules absorbed
= $0.01\times \frac{100}{1000}\times N_A=6\times {10}^{20}$
Total area of charcoal =$2\times 3\times {10}^2$
= $600m^2$
Area per molecule = $\frac{600}{6\times {10}^{20}}$
= $1\times {10}^{-18}{m}^2$
= $1\times {10}^{-18}\times {10}^{4}c{m}^2$
= $1\times {10}^{-14}c{m}^2$