Charcoal is mixed in 500 mL of 0.5 M acetic | vedclass

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(B) Initial moles of acetic acid: $(n_{CH_3COOH})_{	ext{initial}} = 500 \ mL 	imes 0.5 \ M = 250 \ mmol = 0.25 \ mol$.Final moles of acetic acid: $(

Vedclass · Accepted Answer

$6 	imes 10^{22} 	ext{ molecules}$

Vedclass · Answer

(B) Initial moles of acetic acid: $(n_{CH_3COOH})_{	ext{initial}} = 500 \ mL 	imes 0.5 \ M = 250 \ mmol = 0.25 \ mol$.Final moles of acetic acid: $(n_{CH_3COOH})_{	ext{final}} = 500 \ mL 	imes 0.3 \ M = 150 \ mmol = 0.15 \ mol$.Moles adsorbed on charcoal: $\Delta n = 0.25 \ mol - 0.15 \ mol = 0.1 \ mol$.Number of molecules adsorbed: $N = \Delta n 	imes N_A = 0.1 	imes 6 	imes 10^{23} = 6 	imes 10^{22} 	ext{ molecules}$.

Charcoal is mixed in $500 \ mL$ of $0.5 \ M$ acetic acid solution. After some time,the concentration of the solution becomes $0.3 \ M$. Find the total number of molecules adsorbed on the surface of the charcoal.

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