1.

One gram of charcoal adsorbs 100 ml of 0.5M CH_3CO OH & then molarity of acetic acid reduces to 0.49 M. The no. of milli moles of acetic acid adsorbed is _____

Answer»


SOLUTION :`M = (WT)/(GMt) cdot 1000/V , 0.5 (wt)/(60)cdot (1000)/(100)` ,after `0.49 = (wt)/(60) cdot 1000/100`
w.t = 3 , w.t. 2.94 , wt REDUCED = 0.06 , no. of mole = `(0.06)/(60) = 0.001,` Milimoles = 1


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