KVPY Sample Paper KVPY Stream-SX Model Paper-26

3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06 N) 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) 42 mg

B) 54 mg

C) 18mg

D) 36 mg

Correct Answer: C

Solution :

Let the weight of acetic acid initially be \[{{\operatorname{w}}_{1}}\] in 50 mL of 0.060 N solution.

Let the \[\operatorname{N}=\frac{{{w}_{1}}\times 1000}{M.wt.\times 50}\](Normality =0.06 N)

\[0.06=\frac{{{w}_{1}}\times 1000}{60\times 50}\]\[\Rightarrow \]\[{{w}_{1}}=\frac{0.06\times 60\times 50}{1000}=0.18g=180mg\]

After an hour, the strength of acetic acid \[=0.042\,\operatorname{N}\]

So, let the weight of acetic acid be \[{{w}_{2}}\]

\[\operatorname{N}=\frac{{{w}_{2}}\times 1000}{60\times 50};0.042=\frac{{{w}_{2}}\times 1000}{3000}\]\[\Rightarrow \,\,\,{{w}_{2}}=0.126g=126mg\]

Amount of acetic acid adsorbed per g\[=\frac{54}{3}=18mg\]