$3$ 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?

18

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36

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42

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54

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Solution

The correct option is C $18$ mg
Initial millimole of acetic acid solution = N $\times$ Volume = $0.06 N \times 50 = 3$ millimole
Final millimole of acetic acid solution = $0.042 N \times 50 = 2.1$ millimole
Acetic acid adsorbed = 3- 2.1 = 0.9 millimole
Weight after 60 minutes = $\frac{0.9}{10^{3}} \times 60 = 54 m g$
Amount of acetic acid adsorbed per gram of charcoal = $\frac{54}{3} = 18$ mg

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3 g of activated charcoal was added to 50mL of acetic acid solution (0.06N) in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042N. The amount of acetic acid adsorbed(per gram of charcoal) is?

$3$ 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?