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

Given, the chemical reaction

$HCl+{Na}_2{CO}_3⟶NaHCO{}_3+H_{2}O$

$HCl+{NaHCO}_3⟶NaCl+{CO}_2+H_{2}O$

with NaOH the reaction will be

$\mathrm{HCl}+\mathrm{NaOH}⟶\mathrm{NaCl}+{\mathrm{H}}_2\mathrm{O}$

 Let n moles of {NaOH} be x

$\mathrm{And} \mathrm{n} \mathrm{moles} \mathrm{of} {{\mathrm{Na}}_2\mathrm{CO}}_3 \mathrm{be} \mathrm{y}$

First endpoint

$\left(NaOH+{Na}_2{CO}_3\right)=HCl(x+y)\times 1=\frac{1}{10}\times 17.5$

x+y=1.75 -------(i)

Second endpoint

$(NaOH+N{a}_{2}C{O}_3)=HCl$

$x+y\times 2=\frac{1}{10}\times 19$ ---------(ii)

x+2 y=1.9

From (i) and(ii)

Y= 0.15

Weight percentage of ${N{a}_{2}CO}_3=\frac{0.15\times {10}^{-3}\times 106}{0.4}\times 100=3.975\%\cong 4$

Therefore, weight % ${NaCO}_3 is 4.$