2.0 g of activated charcoal is added to 100 | vedclass

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Question

(C) Initial concentration of acetic acid $= 0.5 \ M$. Final concentration of acetic acid $= 0.4 \ M$. Change in concentration $= 0.5 \ M - 0.4 \ M = 0

Vedclass · Accepted Answer

$0.3$

Vedclass · Answer

(C) Initial concentration of acetic acid $= 0.5 \ M$. Final concentration of acetic acid $= 0.4 \ M$. Change in concentration $= 0.5 \ M - 0.4 \ M = 0.1 \ M$. Volume of solution $= 100 \ mL = 0.1 \ L$. Change in moles of acetic acid $= 	ext{Concentration change} 	imes 	ext{Volume in } L = 0.1 \ M 	imes 0.1 \ L = 0.01 \ mol$. Mass of acetic acid adsorbed $= 	ext{Moles} 	imes 	ext{Molar mass} = 0.01 \ mol 	imes 60 \ g \ mol^{-1} = 0.6 \ g$. Mass of charcoal used $= 2.0 \ g$. Mass adsorbed per gram of charcoal $= \frac{0.6 \ g}{2.0 \ g} = 0.3 \ g \ g^{-1}$.

$2.0 \ g$ of activated charcoal is added to $100 \ mL$ of $0.5 \ M$ acetic acid (molar mass $60 \ g \ mol^{-1}$),shaken well and filtered. The concentration of the solution is reduced to $0.4 \ M$. How many grams of acetic acid is adsorbed on charcoal per gram?

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Assertion : $NH_3$ is absorbed more readily over activated charcoal than $CO_2$.
Reason : $NH_3$ is non-polar.

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