Question

# 3 g of activated charcoal was added to 50 ml of acetic acid solution (0.06 N) in a flask. After an hour, it was filtered and the strength of the filtrate was found to be 0.042 N. Amount of acetic acid absorbed (per gram of charcoal) is:A. 54 mgB. 36 mgC. 18 mgD. 42 mg

Hint: This question deals with the volumetric calculations as it involves normality, volume and equivalents. Find the equivalents of acetic acid using the formula $\text{number of equivalents = Normality}\times \text{Volume}\times \text{1}{{\text{0}}^{-3}}$ at initial and final stages and subtract. Then, using an amount of acetic acid by its molecular mass.

Let us solve this question step by step:
Step (1)- Find the number of equivalents of acetic at initial and final timings:
The formula is $\text{number of equivalents = Normality}\times \text{Volume}\times \text{1}{{\text{0}}^{-3}}$; normality at initial stage is 0.06 N and final stage is 0.042 N. The volume of acetic acid is given as 50 ml. The number of equivalents at initial stage are $\text{number of equivalents = 0}\text{.06}\times 50\times \text{1}{{\text{0}}^{-3}}$or 0.003. The number of equivalents at final stage are $\text{number of equivalents = 0}\text{.042}\times 50\times \text{1}{{\text{0}}^{-3}}$or 0.0021.

Step (2)- The number of equivalents of acetic acid consumed or absorbed are the $\text{number of equivalent} {{\text{s}}_{\text{initial}}} - \text{number of equivalent} {{\text{s}}_{\text{final}}}$. So, the equivalents consumed are (0.003-0.0021) is equal to 0.0009 equivalents.

Step (3)- The molecular mass of acetic acid $\left( \text{C}{{\text{H}}_{3}}\text{COOH} \right)$ is
Atomic weight of carbon is 12 grams.
Atomic weight of oxygen is 16 grams.
Atomic weight of hydrogen is 1 gram.
The molar mass will be $\left[ \left( 12\times 2 \right)+\left( 1\times 4 \right)+\left( 16\times 2 \right) \right]$ or 60 grams.

Step (4)- Find the mass of acetic acid absorbed on the piece of charcoal, mass of acetic acid is molar mass of acetic acid multiplied to number of equivalents. So, the mass will be $0.0009\times 60$ or 0.054 grams.

Step (5)- They have asked to find the mass of acetic acid per gram of charcoal and mass of charcoal is given 3 grams. So, the mass of acetic acid per gram will be $\dfrac{0.054}{3}$ or 0.018 grams or 18 milligrams.
The correct answer of this question is 18 mg which is option ‘c’. 18 mg is the amount of acetic acid is absorbed per gram of charcoal.
So, the correct answer is “Option C”.

Note: In this question, we are considering that acetic acid will ionize fully which means complete dissociation. So, the molar mass of acetic acid equals the equivalent mass of acetic acid. As, the tendency of acetic acid to give $\left[ {{\text{H}}^{+}} \right]$ ions is 1. Using the formula, $\text{molarity}\times \text{n-factor=normality}$.