0.4 g mixture of $\mathrm{NaOH},{\mathrm{Na}}_2{\mathrm{CO}}_3$ and some inert impurities was first titrated with $\frac{\mathrm{N}}{10}\mathrm{HCl}$ using phenolphthalein as an indicator, 17.5 mL of HCl was required at the end point. After this methyl orange was added and titrated. 1.5 mL of same HCl was required for the next end point. The weight percentage of ${\mathrm{Na}}_2{\mathrm{CO}}_3$ in the mixture is _________(Rounded off to the nearest integer)

At first end point, gram equivalent of NaOH and Na₂CO₃ is equal to the gram equivalent of HCl.

Assume that moles of NaOH is x and moles of Na₂CO₃ is y.

$$\begin{gathered} \mathrm{gram} \mathrm{equivalent} (\mathrm{NaOH} \mathrm{and} {\mathrm{Na}}_2{\mathrm{CO}}_3)=\mathrm{HCl} \\ \left(\mathrm{x}+\mathrm{y}\right)\times 1=\frac{1}{10}\times 1.75 \\ \mathrm{x}+\mathrm{y}=1.75....\left(1\right) \end{gathered}$$

$\begin{array}{l}\text{At second,end point }\\ \mathrm{gram} \mathrm{equivalents} \mathrm{of} \mathrm{NaOH}+{\mathrm{Na}}_2{\mathrm{CO}}_3=\mathrm{gram} \mathrm{equivalent} \mathrm{of} \mathrm{HCl}\\ \mathrm{x}+\mathrm{y}\times 2=\frac{1}{10}\times 19\\ \mathrm{x}+\mathrm{y}=1.9...\left(2\right)\end{array}$

On solving equations (1) and (2).

y=0.15

$\begin{array}{l}\mathrm{Weight} \%{\mathrm{Na}}_2{\mathrm{CO}}_3=\frac{\mathrm{weight} \mathrm{of} \mathrm{the} \mathrm{compound}}{\mathrm{gram} \mathrm{weight} \mathrm{of} \mathrm{mixture}}\times 100\\ \%{\mathrm{Na}}_2{\mathrm{CO}}_3=\frac{0.15\times {10}^{-3}\times 106}{0.4}\times 100\\ =3.975\%\\ \cong 4\%\end{array}$