$\begin{array}{1 1}(a)\;0.5962g,0.394g&(b)\;0.234g,0.142g\\(c)\;0.1243g,0.235g&(d)\;0.42g,0.24g\end{array}$

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$Na_2CO_3=ag$

$K_2CO_3=bg$

$a+b=1.20$------(1)

For neutralization reaction of 100ml solution.

Meq of $Na_2CO_3$+Meq of $K_2CO_3$=Meq of HCl

$\large\frac{a}{\Large\frac{106}{2}}$$\times 1000+\large\frac{b}{\Large\frac{138}{2}}$$\times 1000=\large\frac{40\times 0.1\times 100}{20}$

$69a+53b=73.14$-------(2)

From (1) & (2)

$a=0.5962g$

$b=0.6038g$

$Na_2CO_3+K_2CO_3$ gives ppt of $BaCO_3$ with $BaCl_2$

Meq of $BaCO_3$=Meq of $Na_2CO_3$+Meq of $K_2CO_3$

$\Rightarrow$ Meq of HCl for 20ml mixture.

$\Rightarrow 40\times 0.1=4$

$\therefore \large\frac{w}{\Large\frac{197}{2}}$$\times =4$

$\therefore$ wt of $BaCO_3=0.394g$

Hence (a) is the correct option.

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