JEE Main & Advanced Chemistry Surface & Nuclear Chemistry / भूतल और नाभिकीय रसायन Question Bank Mock Test - Surface Chemistry

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

A) 15 mg

B) 36 mg

C) 42 mg

D) 54 mg

Correct Answer: A

Solution :

[a] Initial m moles of

\[C{{H}_{3}}COOH=0.06\times 50\]

N=M as basicity of \[C{{H}_{3}}COOH=1\]

Final m moles of \[C{{H}_{3}}COOH=0.042\times 50\]

Hence, mass of \[C{{H}_{3}}COOH\]adsorbed per gram of charcoal

\[=\frac{(0.06-0.042)\times 50\times {{10}^{-3}}\times 60\times {{10}^{3}}}{3}=18mg\]