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3 g of activated charcoal was added to 5...

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

A

18 mg

B

36 mg

C

42 mg

D

54 mg

Text Solution

Verified by Experts

The correct Answer is:
A
Moles of acetic acid initially present
`= (0.06 xx 50)/(1000) = 3 xx 10 ^(-3)`
Moles of acetic acid after adsorption
`= (0.042 xx 50)/(1000) = 2.1 xx 10 ^(-3)`
Moles of acetic acid adsorbed
`= 3.0 xx 10 ^(-3) -2.1 xx 10 ^(-3)`
`= 0.9 xx 10 ^(-3)`
Mass of acetic acid
`=0.9 xx 10 ^(-3) xx 60 = 54 xx 10 ^(-3) g`
Amount of acetic acid adsorbed per gram of charcoal
`= (54 xx 10 ^(-3))/(3) = 18 xx 10 ^(-3) g`
`or =18` mg
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