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 $0.49M$. Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid. Surface area of charcoal $=3.01\times {10}^2{m}^2/gm$.

• A. $5\times {10}^{-19}{m}^2$
• B. $5\times {10}^{+19}{m}^2$
• C. $5\times {10}^{-9}{m}^2$
• D. $5\times {10}^{+9}{m}^2$

Answer 1

The correct option is A $5\times {10}^{-19}{m}^2$

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