Question

$1$ g of charcoal adsorbs $100$ ml of $0.5MC{H}_{3}COOH$ to form a monolayer and the molarity of $C{H}_{3}COOH$ reduces to $0.49$. The surface area of charcoal adsorbed by each molecule of acetic acid is $x\times {10}^{-19}{m}^2$. Surface area of charcoal $=3.01\times {10}^2{m}^2/g$.

Find the value of $x$.

Answer 1

According to given data,

Millimoles of acetic acid taken $=100\times 0.5=50$

Millimoles of acetic acid left $=100\times 0.49=49$

Millimoles of acetic acid adsorbed $=50-49=1$

Molecules of acetic acid adsorbed $=1\times {10}^{-3}\times 6.023\times {10}^{23}$ $=6.023\times {10}^{20}$

Total area of $1$ g charcoal covered by these molecules $=3.01\times {10}^2{m}^2$

$\therefore$ Area covered by 1 molecule $=\frac{3.01\times {10}^2}{6.023\times {10}^{20}}$ $=5\times {10}^{-19}{m}^2$ ($\because$ unilayer adsorption)

Therefore, the value of $x$ is $5$