Solutions

Solutions

Question

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:

  • 18 mg

  • 36 mg

  • 42 mg

  • 54 mg

Answer

A.

18 mg

The initial strength of acetic acid = 0.06N
Final strength = 0.042 N
Volume given = 50 mL
there Initial m moles of CH3COOH 
 = 0.06 x 50 = 3
Final m moles of CH3COOH 
 = 0.042 x 50 = 21
therefore, m moles of CH3COOH absorbed
 = 3-2.1
 = 0.9 m mol
Hence, mass of CH3COOH  absorbed per gram of charcoal
 = fraction numerator 0.9 space straight x space 60 over denominator 3 end fraction
54 over 3 space equals space 18 space mg

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