Question

1 g charcoal adsorbs 100 ml of 0.5 M $C{H}_{3}CHOOH$ to form a monolayer. As a result molarity of acetic acid reduces to 0.49 M. 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. $5.0\times {10}^{-19}{m}^2$
• C. ${10}^{-18}{m}^2$
• D. $2.0\times {10}^{-18}{m}^2$

Answer 1

The correct option is B

$5.0\times {10}^{-19}{m}^2$

Acetic acid adsorbed = 0.5 – 0.49 M = 0.01 M
$\therefore$ Acetic acid adsorbed from 100 mL solution = 0.001 mole
Acetic acid adsorbed by 1 g of charcoal = 0.001 mole $=6.02\times {10}^{20}$molecules.
Surface area of 1 g of charcoal $=3.01\times {10}^2{m}^2$
$\therefore$ Surface area of charcoal covered by each molecule
$=\frac{(3.01\times {10}^2{m}^2)}{(6.02\times {10}^{20}})$
$=5\times {10}^{-19}$