1.

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:

Answer»

18 mg
36 mg
42 mg
54 mg

Solution :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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