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

Given: Mass of oxalic acid dihydrate ($w$) = $6.3 \text{ g}$ Volume of solution ($V$) = $250 \text{ mL} = 0.25 \text{ L}$ Formula of oxalic acid dihydrate is $\text{H}_2\text{C}_2\text{O}_4 \cdot 2\text{H}_2\text{O}$. Molar mass of $\text{H}_2\text{C}_2\text{O}_4 \cdot 2\text{H}_2\text{O} = 2(1) + 2(12) + 4(16) + 2(18) = 2 + 24 + 64 + 36 = 126 \text{ g/mol}$. Since oxalic acid is a dibasic acid, its n-factor is $2$. Equivalent mass of oxalic acid ($E$) = $\frac{\text{Molar mass}}{\text{n-factor}} = \frac{126}{2} = 63 \text{ g/eq}$.

Normality of oxalic acid solution ($N_1$) = $\frac{w}{E \times V \text{ (in L)}} = \frac{6.3}{63 \times 0.25} = \frac{0.1}{0.25} = 0.4 \text{ N}$.

Now, $10 \text{ mL}$ of this solution is neutralized by $0.1 \text{ N NaOH}$. According to the principle of equivalence, for complete neutralization: $N_1 V_1 = N_2 V_2$ Where, $N_1 = 0.4 \text{ N}$ (Normality of oxalic acid) $V_1 = 10 \text{ mL}$ (Volume of oxalic acid) $N_2 = 0.1 \text{ N}$ (Normality of NaOH) $V_2 = ?$ (Volume of NaOH required)

Substituting the values: $0.4 \times 10 = 0.1 \times V_2$ $4 = 0.1 \times V_2$ $V_2 = \frac{4}{0.1} = 40 \text{ mL}$.

Therefore, the required volume of $\text{NaOH}$ is $40 \text{ mL}$.