Question

: One gram of charcoal adsorbs ${\text{100mL}}$ of ${\text{0}}{\text{.5M}}$ ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$ to form a monolayer and thereby the molarity of acetic acid is reduced to ${\text{0}}{\text{.49M}}$ . Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid. Surface area of charcoal $= 3.01\times 10^{2}m^{2}/gm$ .

Answer 1

Hint:The term ‘adsorption’ refers to the attraction and retaining of the molecules of a substance on the surface of a solid or liquid which results in a higher concentration of the molecules on the surface. The substance that is adsorbed is called the adsorbate and the substance on which it is adsorbed is called the adsorbent. It is different from absorption as the former is a surface phenomenon while the latter is a bulk phenomenon.
According to the mole concept, one mole of atoms, ions, molecules is equal to the Avogadro’s number of atoms, ions, molecules respectively. The number $6.023\times 10^{23}$ is called the Avogadro number.

Complete step by step answer:
Calculation of initial amount of acetic acid in $100mL$ of solution.
${\text{1M}}$ acetic acid solution means that 1 mole of acetic acid or 60 gram of acetic acid (since molar mass of acetic acid is equal to 60 gram per mole) is present in ${\text{1000mL}}$ of solution.
Therefore, ${\text{0}}{\text{.5M}}$ acetic acid will mean ${\text{1000mL}}$ of solution contains acetic acid
 $
= 0.5\times 1
= 0.5mole
 $
Therefore, ${\text{100mL}}$ of ${\text{0}}{\text{.5M}}$ solution will contain acetic acid
$
= \dfrac{0.5}{1000}\times 100
= 0.05mole
 $
Therefore, initial moles of acetic acid $= 0.05mole$
Calculation of amount of acetic acid in $100mL$ solution after adsorption.
${\text{1M}}$ acetic acid solution means that 1 mole of acetic acid or 60 gram of acetic acid (since molar mass of acetic acid is equal to 60 gram per mole) is present in ${\text{1000mL}}$ of solution.
Therefore, ${\text{0}}{\text{.49M}}$ acetic acid will mean ${\text{1000mL}}$ of solution contains acetic acid
$
= 0.49\times 1
= 0.49mole
 $

Therefore, ${\text{100mL}}$ of ${\text{0}}{\text{.49M}}$ solution will contain acetic acid
$
= \dfrac{0.49}{1000}\times 100
= 0.049mole
 $
Therefore, final moles of acetic acid $= 0.049mole$
Calculation of amount of acetic acid adsorbed per gram.
Therefore, the number of moles of acetic acid adsorbed per gram of charcoal
$
  {\text{ = 0}}{\text{.05 - 0}}{\text{.049}} \\
  {\text{ = 0}}{\text{.001mole}} \\
 $
According to mole concept, one mole of any species is equal to $6.023\times 10^{23}$ number of particles, therefore the number of molecules of acetic acid adsorbed per gram of charcoal
$
= 0.001\times 6.023\times 10^{23}
= 6.023\times 10^{20}
 $
Calculation of the surface area occupied by each acetic acid molecule.
The given surface area of charcoal is $= 3.01\times 10^{2}m^{2}/gm$.
Therefore, the surface area of the charcoal occupied by each acetic acid molecule
$
= \dfrac{3.01\times 10^{2}}{6.023\times 10^{20}}m^{2}
= 0.499\times 10^{-18}m^{2}
= 5\times 10^{-19}m^{2}
 $

Note:
The nature and surface area of the adsorbent affects the adsorption of gases by solids. The same gas is adsorbed by different solids to different extents at the same temperature. Greater the surface area of the adsorbent, greater will be the volume of the gas adsorbed. This is why charcoal is a very good adsorbent as it has highly porous structure and hence large surface areas.