$\begin{array}{1 1}(a)\;66.25\%,33.75\%&(b)\;60\%,40\%\\(c)\;60.25\%,40.75\%&(d)\;65\%,35\%\end{array}$

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Weight of $Na_2CO_3$='a'g

Weight of $NaCl$='b'g

$a+b=4$----(1)

The solution reacts with HCl,only $Na_2CO_3$ reacts.

$\therefore$ In 25ml,Meq of $Na_2CO_3$=Meq of HCl

$\Rightarrow 50\times \large\frac{1}{10}$

$\Rightarrow 5$

$\therefore$Meq of $Na_2CO_3$ in 250ml=$\large\frac{5\times 250}{25}$

$\Rightarrow 50$

$\large\frac{a}{\Large\frac{106}{2}}$$\times 1000=50$

$a=2.65$

From equ(1)

$b=4-2.65=1.35g$

% of $Na_2CO_3=\large\frac{2.65}{4}$$\times 100=66.25\%$

$\therefore$ % of NaCl=33.75%

Hence (a) is the correct answer.

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