Three grams of activated charcoal was added to 50ml 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. Amount of acetic acid adsorbed (per gram of charcoal) is:

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Solution

Initial millimoles of acetic acid= $N V$
$= 50 \times 0.06$
$= 3 m m o l$
Final millimoles of $C H_{3} C O O H = 50 \times 0.42$
$= 2.1 m m o l$
Acetic acid adsorbed= $3 - 2.1 = 0.9 m m o l$
Weight after $60 m i n = \frac{0.9}{10^{3}} \times 60 = 54 m g$
$∴$ Amount adsorbed per $m g$ of charcoal= $\frac{54}{3} = 18 m g$

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3 g

of activated charcoal was added to

50 m L

of the acetic acid solution

(0.06 N)

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:

3 g of activated charcoal was added to 50 ml of acetic acid solution (0.06 M) in a flask. After an hour it was filtered, and the strength of filtrate was found to be 0.042 M. The mass of acetic acid adsorbed per gram of charcoal is:

3 g

of activated charcoal was added to

50 m L

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 absorbed (per gram of charcoal) is ?

Q. 3

g

of activated charcoal was added to 50

m L

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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Three grams of activated charcoal was added to 50ml 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. Amount of acetic acid adsorbed (per gram of charcoal) is:

Initial millimoles of acetic acid= NV=50×0.06=3mmolFinal millimoles of CH3COOH=50×0.42=2.1mmolAcetic acid adsorbed= 3−2.1=0.9mmolWeight after 60min=0.9103×60=54mg∴ Amount adsorbed per mg of charcoal= 543=18mg