Question

One gm of charcoal adsorbs $CH _{3} COOH$ from $100 ml 0.6 M CH _{3} COOH$ aqueous solution to form a monolayer, and thereby the molarity of $CH _{3} COOH$ reduces to $0.58$. Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid. Surface area of charcoal $=3.0 \times 10^{2} m ^{2} / gm$.

• A. $5 nm ^{2}$
• B. $0.5 nm ^{2}$
• C. $0.25 nm ^{2}$
• D. $2.5 nm ^{2}$

Answer 1

Answer: (C)

mm of $CH _{3} COOH$ adsorbed $=2$
number of molecule $=2 \times 10^{-3} \times 6 \times 10^{23}$
$=12 \times 10^{20}$
area adsorbed by each molecule $=\frac{3 \times 10^{6}}{12 \times 10^{20}}$
$=0.25 \times 10^{-14} cm ^{2}$
$=25 \times 10^{-16} cm ^{2}=0.25 \times 10^{-14} cm ^{2}$
$=0.25 nm ^{2}$