Question

One gram of charcoal adsorbs $400$ mL of $0.5M$ acetic acid to form a monolayer, and the molarity of acetic acid reduces to $0.49M$. Calculate the surface area of charcoal adsorbed by each molecule of acetic acid, where the surface area of charcoal is $3.01\times {10}^2$ m${}^2$ g${}^{-1}$.

• A. $1\times {10}^{-19}{m}^2$
• B. $1.2\times {10}^{-19}{m}^2$
• C. $2.5\times {10}^{-21}{m}^2$
• D. None of these

Answer 1

The correct option is B $1.2\times {10}^{-19}{m}^2$

$400mL$ of $(0.5-0.49=0.01)M$ acetic acid contains $0.4\times 0.01=0.2$ moles of acetic acid or $4\times {10}^{-3}\times 6.023\times {10}^{23}=2.409\times {10}^{21}$ molecules of acetic acid.

Total surface area of 1 g of charcoal is $3.01\times {10}^2{m}^2$.

Thus, the surface area occupied by 1 molecule of acetic acid is $\frac{3.01\times {10}^2{m}^2}{2.409\times {10}^{21}}=1.2\times {10}^{-19}{m}^2$.

Hence, the correct option is $\text{B}$