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(C) The initial amount of acetic acid in $50 \, mL$ of $0.06 \, N$ solution is calculated as:$w_1 = \frac{N 	imes M_{wt} 	imes V}{1000} = \frac{0.06

Vedclass · Accepted Answer

$18$

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(C) The initial amount of acetic acid in $50 \, mL$ of $0.06 \, N$ solution is calculated as:$w_1 = \frac{N 	imes M_{wt} 	imes V}{1000} = \frac{0.06 	imes 60 	imes 50}{1000} = 0.18 \, g = 180 \, mg$.The final amount of acetic acid in the filtrate after adsorption is:$w_2 = \frac{0.042 	imes 60 	imes 50}{1000} = 0.126 \, g = 126 \, mg$.The amount of acetic acid adsorbed by $3 \, g$ of charcoal is:$w_{ads} = 180 \, mg - 126 \, mg = 54 \, mg$.The amount of acetic acid adsorbed per gram of charcoal is:$\frac{54 \, mg}{3 \, g} = 18 \, mg/g$.

$3 \, g$ 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: .............. $mg$

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