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

  • a)
    42 mg

  • b)
    18 mg

  • c)
    54 mg 

  • d)
    36 mg

Correct answer is option 'B'. Can you explain this answer?
Verified Answer
3g of activated charcoal was added to 50 ml of acetic acid solution (0...
To find the amount of acetic acid adsorbed per gram of charcoal, we need to calculate the mass of acetic acid adsorbed by the activated charcoal and divide it by the mass of charcoal used.
The initial concentration of acetic acid is 0.06 M, and after adsorption, the final concentration is 0.042 M. The difference between the initial and final concentrations represents the amount of acetic acid adsorbed by the activated charcoal.
0.06 M - 0.042 M = 0.018 M
Now we need to find the mass of acetic acid (in grams) that was adsorbed by the activated charcoal. We can use the equation: molarity x volume = moles
0.018 M x 50 ml = 0.9 moles
Now, we can find mass of acetic acid adsorbed by mass = moles x molar mass
mass = 0.9 moles x 60 g/mol = 54 g
Now we need to divide the mass of acetic acid adsorbed by the mass of activated charcoal used:
54 g / 3 g = 18 g/g
Therefore, the amount of acetic acid adsorbed per gram of charcoal is 18 mg, So the correct answer is 2.
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Most Upvoted Answer
3g of activated charcoal was added to 50 ml of acetic acid solution (0...
To find the amount of acetic acid adsorbed per gram of charcoal, we need to calculate the mass of acetic acid adsorbed by the activated charcoal and divide it by the mass of charcoal used.
The initial concentration of acetic acid is 0.06 M, and after adsorption, the final concentration is 0.042 M. The difference between the initial and final concentrations represents the amount of acetic acid adsorbed by the activated charcoal.
0.06 M - 0.042 M = 0.018 M
Now we need to find the mass of acetic acid (in grams) that was adsorbed by the activated charcoal. We can use the equation: molarity x volume = moles
0.018 M x 50 ml = 0.9 moles
Now, we can find mass of acetic acid adsorbed by mass = moles x molar mass
mass = 0.9 moles x 60 g/mol = 54 g
Now we need to divide the mass of acetic acid adsorbed by the mass of activated charcoal used:
54 g / 3 g = 18 g/g
Therefore, the amount of acetic acid adsorbed per gram of charcoal is 18 mg, So the correct answer is 2.
Free Test
Community Answer
3g of activated charcoal was added to 50 ml of acetic acid solution (0...
Molarity of CH3COOH adsorbed
= 0.06 – 0.042 = 0.018
Number of milli moles of CH3COOH that are
adsorbed = 0.018 × 50 ml = 0.90
Final Amount = 0.90 × 60 = 54 mg
[∵ molar mass of CH3COOH is taken as 60 g mol-1]
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3g of activated charcoal was added to 50 ml of acetic acid solution (0.06 M) in a flask. After an hour it was filtered and the strength of filtrate was found to be 0.042 M. The amount of acetic acid adsorbed per gram of charcoal isa)42 mgb)18 mgc)54 mgd)36 mgCorrect answer is option 'B'. Can you explain this answer?
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3g of activated charcoal was added to 50 ml of acetic acid solution (0.06 M) in a flask. After an hour it was filtered and the strength of filtrate was found to be 0.042 M. The amount of acetic acid adsorbed per gram of charcoal isa)42 mgb)18 mgc)54 mgd)36 mgCorrect answer is option 'B'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about 3g of activated charcoal was added to 50 ml of acetic acid solution (0.06 M) in a flask. After an hour it was filtered and the strength of filtrate was found to be 0.042 M. The amount of acetic acid adsorbed per gram of charcoal isa)42 mgb)18 mgc)54 mgd)36 mgCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 3g of activated charcoal was added to 50 ml of acetic acid solution (0.06 M) in a flask. After an hour it was filtered and the strength of filtrate was found to be 0.042 M. The amount of acetic acid adsorbed per gram of charcoal isa)42 mgb)18 mgc)54 mgd)36 mgCorrect answer is option 'B'. Can you explain this answer?.
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