When 200 , mL of 0.2 , M acetic acid is shake | vedclass

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Question

(C) Initial moles of acetic acid $= M_1 	imes V_1 = 0.2 \, M 	imes 0.2 \, L = 0.04 \, mol$. Final moles of acetic acid $= M_2 	imes V_2 = 0.1 \, M

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$2$

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(C) Initial moles of acetic acid $= M_1 	imes V_1 = 0.2 \, M 	imes 0.2 \, L = 0.04 \, mol$. Final moles of acetic acid $= M_2 	imes V_2 = 0.1 \, M 	imes 0.2 \, L = 0.02 \, mol$. Moles of acetic acid adsorbed $= 0.04 - 0.02 = 0.02 \, mol$. Molar mass of acetic acid $(CH_3COOH) = 60 \, g/mol$. Mass of acetic acid adsorbed $= 0.02 \, mol 	imes 60 \, g/mol = 1.2 \, g$. Mass of wood charcoal $= 0.6 \, g$. Mass of acetic acid adsorbed per gram of carbon $= \frac{1.2 \, g}{0.6 \, g} = 2 \, g/g$.

When $200 \, mL$ of $0.2 \, M$ acetic acid is shaken with $0.6 \, g$ of wood charcoal,the final concentration of acetic acid after adsorption is $0.1 \, M$. The mass of acetic acid adsorbed per gram of carbon is $...... \, g$.

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