An aqueous solution of 18 g oxalic acid (H_2 | vedclass

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(A) The molar mass of oxalic acid dihydrate $(H_2C_2O_4 \cdot 2H_2O)$ is $126 \ g/mol$. Moles of oxalic acid in $400 \ mL$ = $\frac{18 \ g}{126 \ g/mo

Vedclass · Accepted Answer

$500$

Vedclass · Answer

(A) The molar mass of oxalic acid dihydrate $(H_2C_2O_4 \cdot 2H_2O)$ is $126 \ g/mol$. Moles of oxalic acid in $400 \ mL$ = $\frac{18 \ g}{126 \ g/mol} = \frac{1}{7} \ mol$. Moles of oxalic acid in $50 \ mL$ = $\frac{1}{7} 	imes \frac{50}{400} = \frac{1}{56} \ mol$. The neutralization reaction is: $H_2C_2O_4 + 2NaOH ightarrow Na_2C_2O_4 + 2H_2O$. According to the stoichiometry,$1 \ mol$ of $H_2C_2O_4$ reacts with $2 \ mol$ of $NaOH$. Moles of $NaOH$ required = $2 	imes \frac{1}{56} = \frac{1}{28} \ mol$. Volume of $0.1 \ M \ NaOH$ = $\frac{	ext{moles}}{	ext{molarity}} = \frac{1/28}{0.1} = \frac{10}{28} \approx 0.357 \ L = 357 \ mL$. Note: If anhydrous oxalic acid ($H_2C_2O_4$,molar mass $90 \ g/mol$) is assumed: Moles in $400 \ mL$ = $\frac{18}{90} = 0.2 \ mol$. Moles in $50 \ mL$ = $0.2 	imes \frac{50}{400} = 0.025 \ mol$. Moles of $NaOH$ required = $2 	imes 0.025 = 0.05 \ mol$. Volume of $0.1 \ M \ NaOH$ = $\frac{0.05}{0.1} = 0.5 \ L = 500 \ mL$. Given the options,the calculation assumes anhydrous oxalic acid.

An aqueous solution of $18 \ g$ oxalic acid $(H_2C_2O_4 \cdot 2H_2O)$ is made up to $400 \ mL$. Calculate the volume of $0.1 \ M \ NaOH$ required to completely neutralize $50 \ mL$ of the above solution in $mL$.

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