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

The correct option is D

$40mL$

The explanation for the correct option

(D) $\mathbf{40}\mathbf{}\mathit{m}\mathit{L}$

Step 1:Normality of oxalic acid dihydrate ${\mathbf{H}}_{\mathbf{2}}{\mathbf{C}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}\mathbf{.}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$

• Normality is the gram equivalent of a solute present per liter of a given solution. It can be calculated as follows,

Normality $=$Molarity$\times$n-factor

• Here the given solute is oxalic acid dihydrate. Oxalic acid is a dibasic acid because there are 2 protons that can be replaced. Therefore its n factor will be 2.
• Molarity $=\frac{Numberofmolesofsolute}{Volumeofsolutoininliter}$
• Number of moles of oxalic acid (molar mass of oxalic acid is $126gmo{l}^{-1}$),

$$\begin{gathered} =\frac{givenmass}{molarmass} \\ =\frac{6.3}{126} \\ =0.05mol \end{gathered}$$

• Therefore,

$$\begin{gathered} Molarity=\frac{0.05}{250}\times 1000 \\ =0.2M \end{gathered}$$

• Note that the given value of the solution is $250mL$, In order to convert $250mL$to liter we need to divide $250$ by $1000$. $(1L=1000mL)$
• Thus normality will be equal to,

$$\begin{gathered} =Molarity\times nfactor \\ =0.2\times 2 \\ =0.4N \end{gathered}$$

Step 2: Volume of sodium hydroxide( $\mathit{N}\mathit{a}\mathit{O}\mathit{H}$ ) solution

• Apply the above-founded values in the normality equation;

$N_1{V}_1=N_2{V}_2$

Where,

$N_1=$Normality of oxalic acid

$N_2=$Normality of $\mathrm{NaOH}$

$V_1=$Volume of oxalic acid taken for neutralization

$V_2=$Volume of $\mathrm{NaOH}$ (need to find).

• So that we get,

$$\begin{gathered} 0.4N\times 10mL=0.1N\times V_2 \\ {V}_2=\frac{0.4N\times 10mL}{0.1N} \\ =40mL \end{gathered}$$

Explanation for the incorrect options

Since, the required volume is $40mL$. Therefore, options (A), (B), and (C) stand incorrect.

Therefore the correct option is (D) $\mathbf{40}\mathbf{}\mathit{m}\mathit{L}$.