Verified Solution • 2025–26 Session moderate

QJEEM22Ch9583: JEE Main 2022 Chemistry - Class 12 JEEM22Ch9583 Solution: 100 mL of 0.3 M acetic acid is shaken with 0.8 g of wood…

100 mL of 0.3 M acetic acid is shaken with 0.8 g of wood charcoal. The final concentration of acetic acid in the solution after adsorption is 0.125 M. The mass of acetic acid adsorbed per gram of carbon is _____________ ×\times 10 −- 4 g.

(Given : Molar mass of acetic acid = 60 g mol −- 1)

Step-by-Step Verified Solution

amp; Initial moles of acetic acid =M1×V1amp;=100×0.3=30 m mol.\begin{aligned}&\text { Initial moles of acetic acid }=M_{1} \times V_{1} \\\\&=100 \times 0.3=30 \mathrm{~m} \mathrm{~mol} .\end{aligned}

amp; moles after adsorption =100×0.125=12.5 m mol. amp; moles adsorbed =30−12.5=17.5 m mol\begin{aligned}&\text { moles after adsorption }=100 \times 0.125=12.5 \mathrm{~m} \mathrm{~mol} \text {. } \\\\&\text { moles adsorbed }=30-12.5=17.5 \mathrm{~m} \mathrm{~mol}\end{aligned}

for acetic acid; weight adsorbed.

amp; moles = Weight  Molar Mass amp; weight =17.5×10−3×60=1050×10−3 gamp; for 0.8 g charcoal mass of acetic acid adsorbed amp;=1050×10−3 gamp;1 g charcoal =1050×10−30.8=1312.5×10−3 gamp;=13125×10−4 g\begin{aligned}&\text { moles }=\frac{\text { Weight }}{\text { Molar Mass }} \\\\&\text { weight }=17.5 \times 10^{-3} \times 60=1050 \times 10^{-3} \mathrm{~g} \\\\&\text { for } 0.8 \mathrm{~g} \text { charcoal mass of acetic acid adsorbed } \\\\&=1050 \times 10^{-3} \mathrm{~g} \\\\&1 \mathrm{~g} \text { charcoal }=\frac{1050 \times 10^{-3}}{0.8}=1312.5 \times 10^{-3} \mathrm{~g} \\\\&=13125 \times 10^{-4} \mathrm{~g}\end{aligned}

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