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Question

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:

A
18 mg
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B
36 mg
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C
42 mg
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D
54 mg
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Solution

The correct option is A 18 mg
50 mLof 0.06 N acetic acid solution contains 0.06 mol/L×50 mL1000 mL/L×60 g/mol=0.180 gof acetic acid.
After filtration, 50 mL of 0.042 N acetic acid solution contains 0.042mol×50mL1000mL/L=0.126 g of acetic acid.
Hence, the amount of acetic acid adsorbed on 3 g of activated charcoal is 0.180g0.126g=0.054 g. This is equal to 54 mg.
Hence, the amount of acetic acid adsorbed on I g of activated charcoal is 54mg×1g3g= 18 mg.

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