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Question

3 g of activated charcoal was added to 50 mL of the 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 :

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 mL of 0.06 N acetic acid solution contains 0.06 mol/L ×50mL1000mL/L×60g/mol=0.180 g of acetic acid.

After filtration, 50 mL of 0.042 N acetic acid solution contains 0.042mol/L×50 mL1000 mL/L=0.126 g of acetic acid.

Hence, the amount of acetic acid adsorbed on 3 g of activated charcoal is 0.180g0.126g=0.054g. This is equal to 54 mg.

Hence, the amount of acetic acid adsorbed on 1 g of activated charcoal is 54mg×1g3g=18mg.

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