0 . 4   g mixture of NaOH , Na 2 CO 3 and some inert impurities was first titrated with N 10 HCl using…

Upto first end point gm equi. of $\left(\mathrm{NaOH}+{\mathrm{Na}}_2{\mathrm{CO}}_3\right)=\mathrm{HCl}$

$\mathrm{x}+\mathrm{y}\times 1=\frac{1}{10}\times 17.5$

$\mathrm{x}+\mathrm{y}=1.75 ....\left(1\right)$

Upto second end point

$\mathrm{NaOH}+{\mathrm{Na}}_2{\mathrm{CO}}_3\equiv \mathrm{HCl}$

$\mathrm{x}+\mathrm{y}\times 2=\frac{1}{10}\times 19$

$\mathrm{x}+2 \mathrm{y}=1.9 ....\left(2\right)$

$\mathrm{y}=0.15$

$\%{\mathrm{Na}}_2{\mathrm{CO}}_3=\frac{0.15\times {10}^{-3}\times 106}{0.4}\times 100$

$=3.975\%$

$=4\%$

Hence answer is $\left(4\right)$