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Q.

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

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a

18 mg

b

36 mg

c

42 mg

d

54 mg

answer is D.

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Detailed Solution

50 mL of 0.06 N CH3COOH=50×0.06 milliequivalents 50 mL of 0.042 N CH3COOH=50×0.042 milliequivalents CH3COOH adsorbed =50(0.06−0.042) milliequivalent =50×0.0181000 equivalent =50×0.018×601000gCH3COOH=0.054g=0.054×1000mg=54mg

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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