Question

$1g$ charcoal adsorbs $100mL$ of $0.5M$ $C{H}_{3}COOH$ to form a monolayer.As a result molarity of acetic acid reduces to $0.49M$. what will be the surface area covered by each molecule of acetic acid ?Given that surface area of charcoal $=3.01\times {10}^2{m}^2/g$

• A. $2.5\times {10}^{-19}{m}^2$
• B. $5\times {10}^{-19}{m}^2$
• C. ${10}^{-18}{m}^2$
• D. $2.0\times {10}^{-18}{m}^2$

Answer 1

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

Initial $m$ moles $=100\times 0.5=50$

Final $m$ moles $=100\times 0.49=49$

$\therefore$ $m$ moles of $C{H}_{3}COOH$ adsorbed $=1={10}^{-3}$ moles.

No. of molecules adsorbed $=6.02\times {10}^{20}$

Area occupied by each molecule $=\frac{3.01\times {10}^2}{6.02\times {10}^{20}}=5\times {10}^{-19}{m}^2$