50 mL of 1 M oxalic acid is shaken with 0.5 | vedclass

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Question

(C) The initial moles of oxalic acid $= M_1 	imes V = 1 \ M 	imes 0.05 \ L = 0.05 \ mol$.The final moles of oxalic acid $= M_2 	imes V = 0.5 \ M

Vedclass · Accepted Answer

$6.30$

Vedclass · Answer

(C) The initial moles of oxalic acid $= M_1 	imes V = 1 \ M 	imes 0.05 \ L = 0.05 \ mol$.The final moles of oxalic acid $= M_2 	imes V = 0.5 \ M 	imes 0.05 \ L = 0.025 \ mol$.The moles of oxalic acid adsorbed $= 0.05 - 0.025 = 0.025 \ mol$.The molar mass of oxalic acid $(H_2C_2O_4 \cdot 2H_2O)$ is $126 \ g/mol$.The mass of oxalic acid adsorbed by $0.5 \ g$ of charcoal $= 0.025 \ mol 	imes 126 \ g/mol = 3.15 \ g$.Therefore,the amount of oxalic acid adsorbed per $g$ of charcoal $= \frac{3.15 \ g}{0.5 \ g} = 6.30 \ g/g$.

$50 \ mL$ of $1 \ M$ oxalic acid is shaken with $0.5 \ g$ of wood charcoal. The final concentration of the solution after adsorption is $0.5 \ M$. The amount of oxalic acid adsorbed per $g$ of charcoal is ....... $g$.

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