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One gram of charcoal adsorbs 100 ml of 0...

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 _____

Text Solution

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The correct Answer is:
1
`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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Knowledge Check

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