One litre of 0.4 M acetic acid solution is passed through twelve grams of activated animal charcoal. The concentration of acetic acid solution coming out of charcoal was found to be 0.25 M. Value of x/m is (x= weight of adsorbate, ,m= weight of adsorbent; assume no volume changes during the process)
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a
0.75
b
0.5
c
1.5
d
0.15
answer is A.
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Detailed Solution
I) Number of moles of CH3COOH before adsorbtion0.4=n1×11 n1=0.4;II) Number of moles of CH3COOH after adsorption0.25=n2×11 n2=0.25Number of moles of CH3COOH adsorbed = (0.4 - 0.25) = 0.15Weight of CH3COOH adsorbed (x) = 0.15 x 60 =9 g Weight of charcoal (m) = 12 g∴xm=912=0.75