One litre of 0.2 M acetic acid solution is passed through four grams of activated animal charcoal. The concentration of acetic acid solution coming out of charcoal was found to be 0.15 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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Detailed Solution

I) Number of moles of CH3COOH before adsorbtion0.2=n1×11 n1=0.2;II) Number of moles of CH3COOH after adsorption0.15=n2×11 n2=0.15Number of moles of CH3COOH adsorbed = (0.2 - 0.15) = 0.05Weight of CH3COOH adsorbed (x) = 0.05 x 60 =3 gGiven that the weight of charcoal (m) = 4 g∴xm=34=0.75

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