wiz-icon
MyQuestionIcon
MyQuestionIcon
12
You visited us 12 times! Enjoying our articles? Unlock Full Access!
Question

One gram of charcoal adsorbs 100 mL of 0.5 M CH3COOH to form a mono-layer and thereby the molarity of acetic acid is reduced to 0.49 M. The surface area (in m2) of the charcoal adsorbed by each molecule of acetic acid is x×1019 m2. Surface area of charcoal= 3.0115×102 m2/g. Find the value of x
(Given NA=6.023×1023)

Open in App
Solution

Number of moles of acetic acid initially present=MV1000=0.5×1001000=0.05
Number of moles of acetic acid left =MV1000=0.49×1001000=0.049
Number of moles of acetic acid adsorbed=0.050.049=0.001 mol
Number of molecules of acid adsorbed=0.001×6.023×1023=6.023×1020
Area occupied by single molecule of acetic acid=Total areaNumber of molecules adsorbed
=3.01×1026.023×1020
=5×1019 m2

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
All About Adsorption
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon