$(a)\;91.2\%\qquad(b)\;93.4\%\qquad(c)\;90.1\%\qquad(d)\;92.3\%$

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Meq of HCl added to 50ml $Na_2CO_3$ solution =$50\times 0.1=5$

Meq of HCl used for $Na_2CO_3$ in 50ml solution = 5 -1.6 = 3.4

$\therefore$ Meq of HCl used for $Na_2CO_3$ in 250ml solution = $\large\frac{3.4\times 250}{50}$$=17$

$\therefore$ Meq of $Na_2CO_3=17$

$\Rightarrow\large\frac{w}{\Large\frac{106}{2}}$$\times 1000=17$

$\Rightarrow w=0.901g$

$\%$ of $Na_2CO_3$ in the sample = $\large\frac{0.901}{1}$$\times 100=90.1$ %

Hence (c) is the correct answer.

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