Question

One gram of charcoal adsorbs 100mL of 0.5M CH_3COOH 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 × 10⁻¹⁹ m²
• B. 2.5 × 10⁻¹⁹ m²
• C. 5 × 10⁻⁸ m²
• D. 2.5 × 10⁻⁸ m²

Answer 1

The correct option is A

5 × 10⁻¹⁹ m²

$\text{No. of moles of}C{H}_{3}COOH\text{initially present}\frac{MV}{1000}=\frac{0.5\times 100}{1000}=0.05\text{No. of moles of}C{H}_{3}COOH\text{left}\frac{MV}{1000}=\frac{0.49\times 100}{1000}=0.049\text{No. of moles of}C{H}_{3}COOH\text{adsorded}0.05-0.049=0.001mol\text{No. of molecules of acid adsorbed}=0.001\times 6.023\times {10}^{23}=6.023\times 1020\text{Area occupied by single molecules}=\frac{Totalarea}{Numberofmoleculesadsorbed}=\frac{3.01\times {10}^2}{6.023\times {10}^{20}}=5\times {10}^{-19}{m}^2$