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 so be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is

Given, initial strength of acetic acid = 0.06 N

Final strength = 0.042N, volume given = 50 mL

$\therefore$$\text{ Initial millimoles of }{\mathrm{CH}}_3\mathrm{COOH}=0.06\times 50=3$

$\text{ Final millimoles of }{\mathrm{CH}}_3\mathrm{COOH}=0.042\times 50=2.1$

$\therefore \text{ Millimoles of }{\mathrm{CH}}_3\mathrm{COOH}\text{ adsorbed }=3-2.1=0.9\mathrm{mmol}$

                                                            = 0.9x60 mg

                                                            = 54 mg