Question

$5$ mL of $0.3$ M acetic acid is shaken with $5$ g of activated charcoal. The concentration of acetic acid is reduced to $\frac{1}{3}$ of the original molarity. The weight of acetic acid adsorbed per gram of charcoal is:

• A. $2\times {10}^{4}g$
• B. $1.2\times {10}^{-2}g$
• C. $2\times {10}^{-2}g$
• D. $3\times {10}^{-2}g$

Answer 1

The correct option is B $1.2\times {10}^{-2}g$

$5$ mL of $0.3$ M acetic acid is shaken with $5$ g of activated charcoal.

The concentration of acetic acid is reduced to $\frac{1}{3}$ of the original molarity.

Initial millimoles $=5\times 0.3=1.5$

Millimoles left after adsorption $=5\times 0.3\times \frac{1}{3}=0.5$

$\therefore$ Millimoles of acid adsorbed $=1.5-0.5=1.0$

$\therefore$ Moles of acid adsorbed $=1.0\times {10}^{-3}$

Moles adsorbed per g $=\frac{1.0\times {10}^{-3}}{5}=2\times {10}^{-4}$

Weight adsorbed per g $=2\times {10}^{-4}\times 60=0.012g$

Hence, option B is the correct answer.