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. The amount of acetic acid adsorbed (per gram of charcoal) is:

We have a solution of volume $$V_1 = 50\ \text{mL} = 0.050\ \text{L}$$ whose normality before contact with charcoal is $$N_1 = 0.06\ \text N$$.

For any solution, the number of equivalents is given by the relation $$\text{equivalents} = N \times V(\text{in L}).$$ Substituting the initial data, the initial number of equivalents of acetic acid is

$$\text{Eq}_\text{initial} \;=\; N_1 \times V_1 \;=\; 0.06 \times 0.050 \;=\; 0.003\ \text{eq}.$$

After adsorption, the filtrate still has the same volume $$V_2 = 0.050\ \text{L}$$ but its normality has fallen to $$N_2 = 0.042\ \text N.$$ Again applying the same formula, the remaining equivalents of acetic acid are

$$\text{Eq}_\text{final} \;=\; N_2 \times V_2 \;=\; 0.042 \times 0.050 \;=\; 0.0021\ \text{eq}.$$

The difference between the initial and final equivalents equals the equivalents that have been removed from solution and therefore adsorbed on charcoal:

$$\text{Eq}_\text{adsorbed} \;=\; \text{Eq}_\text{initial} - \text{Eq}_\text{final} \;=\; 0.003 - 0.0021 \;=\; 0.0009\ \text{eq}.$$

Acetic acid $$CH_3COOH$$ is monoprotic, so one equivalent is the same as one mole. Its molar (and hence equivalent) mass is

$$M = 12 \times 2 + 1 \times 4 + 16 \times 2 = 60\ \text{g\ per\ eq}.$$

The mass corresponding to the adsorbed equivalents is therefore

$$m_\text{adsorbed} \;=\; \text{Eq}_\text{adsorbed} \times M \;=\; 0.0009 \times 60 \;=\; 0.054\ \text{g}.$$

Converting grams to milligrams, we obtain

$$0.054\ \text{g} \;=\; 0.054 \times 1000 \;=\; 54\ \text{mg}.$$

This mass of acetic acid has been adsorbed by $$3\ \text{g}$$ of charcoal. The quantity adsorbed per gram of charcoal is therefore

$$\frac{54\ \text{mg}}{3\ \text{g}} \;=\; 18\ \text{mg per g}.$$

Hence, the correct answer is Option B.