Verified Solution • 2025–26 Session moderate

QJEEM15Che973: JEE Main 2015 Chemistry - Class 12 JEEM15Che973 Solution: 3 g of activated charcoal was added to 50 mL of acetic acid…

3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06N) in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is:

Multiple Choice Options

  • A 36 mg
  • B 42 mg
  • C 54 mg
  • D 18 mg

Step-by-Step Verified Solution

Let the weight of acetic acid initially be

w1{w_1} in 5050 mlml of 0.0600.060 NN solution.

Let the N=w1×1000M.wt.×50   N = {{{w_1} \times 1000} \over {M.wt. \times 50}}\,\,\,

(Normality =0.06=0.06 NN )

0.06=w1×100060×500.06 = {{{w_1} \times 1000} \over {60 \times 50}}

⇒     w1=0.06×60×501000\Rightarrow \,\,\,\,\,{w_1} = {{0.06 \times 60 \times 50} \over {1000}}

=0.18 g=180 mg.= 0.18\,g = 180\,mg.

After an hour, the strength of acetic acid

=0.042=0.042 NN

so, let the weight of acetic acid be w2{w_2}

N=w2×100060×50;0.042=w2×10003000N = {{{w_2} \times 1000} \over {60 \times 50}};0.042 = {{{w_2} \times 1000} \over {3000}}

⇒          w2=0.126g=126mg\Rightarrow \,\,\,\,\,\,\,\,\,\,{w_2} = 0.126g = 126mg

So amount of acetic acid adsorbed per 3g3g

=180−126 mg=54 mg= 180 - 126\,mg = 54\,mg

Amount of acetic acid adsorbed per gg

=543=18 mg= {{54} \over 3} = 18\,mg

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