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JEE Main 20154 April
Exam
JEE Main
Year
2015
Date
4 April
Subject
Chemistry
Chapter
Surface Chemistry

JEE Main 2015 · 4 April · Chemistry · Surface Chemistry

3g of activated charcoal was added to 50mL of acetic acid solution...

3g3g of activated charcoal was added to 50mL50mL of acetic acid solution (0.06N)( 0 . 06N ) in a flask. After an hour, it was filtered and the strength of the filtrate was found to be 0.042N0 . 042N. The amount of acetic acid adsorbed (per gram of charcoal) is:

Options

  1. A54mg54mg
  2. B18mg18mg
  3. C36mg36mg
  4. D42mg42mg

Correct answer

B. 18mg18mg

Step-by-step solution

Milliequivalents (Meqs)( {M}_{eqs} ) of CH3COOH{\text{CH}}_{3}\text{COOH} (initial)=50×0.06=3= 50 \times \text{0.06} = \text{3}

Milliequivalents of CH3COOH{\text{CH}}_{3}\text{COOH} (final)=50×0.042=2.1= 50 \times \text{0.042} = \text{2.1}

CH3COOH{\text{CH}}_{3}\text{COOH} adsorbed=3−2.1=0.9 Meqs= 3 - \text{2.1} = {\text{0.9 M}}_{eqs}

=9×10−1×60 g/mol×10−3g= 9 \times 1{0}^{-1} \times \text{60 g} / mol \times 1{0}^{- 3} g

=540×10−4=0.054 g= 5 4 0 \times 1 {0}^{- 4} = \text{0.054 g}

=54mg= 54 mg

Amount of acetic acid adsorbed per gram of charcoal=543=18 mg/g= \dfrac{54}{3} = \text{18 mg/g} of charcoal.
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