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\[=\frac{54}{3}=18mg\] |
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