Q6 Experiment10: Adsorption of Acetic Acid on Charcoal To study the adsorption of charcoal in Acetic acid (standardized 0.40M), charcoal was added to bottles containing different volumes of acetic acid, the mixture was then shaked, filtrated and titrated against 0.1M NaOH solution. The following data was obtained: Mass of C Acetic acid 0.400 M H:0 vol of solution taken for 0.10 M NAOH used added 50 ml. 40 mL titration after filtration 25.0 ml 25.0 mL for titration 2.70 ml. 14.60 mL added 3.00 g 3.10 g 10 1. Find the number of moles of acetic acid adsorbed per gram of C.

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Q6 Experiment10: Adsorption of Acetic Acid on Charcoal
To study the adsorption of charcoal in Acetic acid (standardized 0.40M), charcoal
was added to bottles containing different volumes of acetic acid, the mixture was
then shaked, filtrated and titrated against 0.1M NaOH solution. The following
data was obtained:
Mass of CAcetic acid 0.400 M
H2O vol of solution taken for 0.10 M NAOH used
titration after filtration
25.0 ml.
25.0 mL
added
50 ml.
for titration
2.70 ml.
added
3.00 g
3.10 g
10
40 mL
14.60 mL
1. Find the number of moles of acetic acid adsorbed per gram of C.
2. Given
Nm KNm
Plotted vs C following equation is obtained Y = 96.887 x + 144.95
N
The K will be
a). 0.668
b). 0.868
c). 0.886
d). 0.968
e). 0.896
3. Given Frendlich equation
log N = log K + a log C
Plotted log N vs log C; equation obtained is Y = 0.9106 x- 2.2772
Find "K"
a). 5.28 x 104
b). 2.85 x 104
c). 5.28 x 103
d). 8.25 x 103
e). 9.11 x 104
12
Transcribed Image Text:Q6 Experiment10: Adsorption of Acetic Acid on Charcoal To study the adsorption of charcoal in Acetic acid (standardized 0.40M), charcoal was added to bottles containing different volumes of acetic acid, the mixture was then shaked, filtrated and titrated against 0.1M NaOH solution. The following data was obtained: Mass of CAcetic acid 0.400 M H2O vol of solution taken for 0.10 M NAOH used titration after filtration 25.0 ml. 25.0 mL added 50 ml. for titration 2.70 ml. added 3.00 g 3.10 g 10 40 mL 14.60 mL 1. Find the number of moles of acetic acid adsorbed per gram of C. 2. Given Nm KNm Plotted vs C following equation is obtained Y = 96.887 x + 144.95 N The K will be a). 0.668 b). 0.868 c). 0.886 d). 0.968 e). 0.896 3. Given Frendlich equation log N = log K + a log C Plotted log N vs log C; equation obtained is Y = 0.9106 x- 2.2772 Find "K" a). 5.28 x 104 b). 2.85 x 104 c). 5.28 x 103 d). 8.25 x 103 e). 9.11 x 104 12
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