$3 g$ of activated charcoal was added to $50 m L$ of the acetic acid solution $(0.06 N)$ in a flask after an hour, it 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:

18 m g

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36 m g

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42 m g

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54 m g

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Solution

The correct option is A $18 m g$
50 mLof 0.06 N acetic acid solution contains $0.06 m o l / L \times \frac{50 m L}{1000 m L / L} \times 60 g / m o l = 0.180 g$ of acetic acid.
After filtration, $50 m L$ of 0.042 N acetic acid solution contains $0.042 m o l \times \frac{50 m L}{1000 m L / L} = 0.126 g$ of acetic acid.
Hence, the amount of acetic acid adsorbed on 3 g of activated charcoal is $0.180 g — 0.126 g = 0.054 g$ . This is equal to $54 m g$ .
Hence, the amount of acetic acid adsorbed on I g of activated charcoal is $\frac{54 m g \times 1 g}{3 g} = 18 m g$ .

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3 g of activated charcoal was added to 50 mL of the acetic acid solution (0.06N) in a flask after an hour, it 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:

$3 g$ of activated charcoal was added to $50 m L$ of the acetic acid solution $(0.06 N)$ in a flask after an hour, it 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: