An aqueous solution of 6.3 g of oxalic acid | vedclass

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Question

(A) The molar mass of oxalic acid dihydrate $(H_2C_2O_4 \cdot 2H_2O)$ is $126 \ g/mol$. Since it is a dibasic acid,its equivalent mass is $126 / 2 = 6

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(A) The molar mass of oxalic acid dihydrate $(H_2C_2O_4 \cdot 2H_2O)$ is $126 \ g/mol$. Since it is a dibasic acid,its equivalent mass is $126 / 2 = 63 \ g/eq$. First,calculate the normality $(N_1)$ of the oxalic acid solution: $N_1 = \frac{	ext{mass}}{	ext{equivalent mass}} 	imes \frac{1000}{V(mL)} = \frac{6.3}{63} 	imes \frac{1000}{250} = 0.1 	imes 4 = 0.4 \ N$. Using the law of equivalence for neutralization: $N_1 V_1 = N_2 V_2$ $(0.4 \ N) 	imes (10 \ mL) = (0.1 \ N) 	imes V_2$ $V_2 = \frac{0.4 	imes 10}{0.1} = 40 \ mL$.

An aqueous solution of $6.3 \ g$ of oxalic acid dihydrate is made up to $250 \ mL$. The volume of $0.1 \ N \ NaOH$ required to completely neutralize $10 \ mL$ of this solution is ............. $mL$.

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