To form a complete monolayer of acetic acid on $1 \mathrm{~g}$ of charcoal, $100 \mathrm{~mL}$ of $0.5…
To form a complete monolayer of acetic acid on $1 \mathrm{~g}$ of charcoal, $100 \mathrm{~mL}$ of $0.5 \mathrm{M}$ acetic acid was used. Some of the acetic acid remained unadsorbed. To neutralize the unadsorbed acetic acid, 40 $\mathrm{mL}$ of $1 \mathrm{M} \mathrm{NaOH}$ solution was required. If each molecule of acetic acid occupies $\mathbf{P} \times 10^{-23} \mathrm{~m}^2$ surface area on charcoal, the value of $\mathbf{P}$ is _______ [Use given data: Surface area of charcoal $=1.5 \times 10^2 \mathrm{~m}^2 \mathrm{~g}^{-1} ;$ Avogadro's number $\left(\mathrm{N}_{\mathrm{A}}\right)=6.0 \times 10^{23}$ $\left.\mathrm{mol}^{-1}\right]$
Solution
$\begin{aligned} & \text { Millimole of acid taken }=100 \times 0.5=50 \\ & \text { Millimole of } \mathrm{NaOH} \text { used }=40 \times 1=40 \\ & \text { Millimole of acid adsorbed }=50-40=10 \\ & \text { Molecules of acid adsorbed }=10 \times 10^{-3} \times 6 \times 10^{23}=6 \times 10^{21} \\ & \text { Surface area occupied per molecule }=\frac{1.5 \times 10^2}{6 \times 10^{21}}=0.25 \times 10^{-19}=2500 \times 10^{-23}\end{aligned}$