$(a)\;10.5g/l,9.5g/l\qquad(b)\;11.05g/l,8.76g/l\qquad(c)\;10.25g/l,10g/l\qquad(d)\;12g/l,7g/l$

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Caustic soda reacts with both HCl and $H_2SO_4$ whereas $BaCl_2$ reacts with only $H_2SO_4$ to precipitate out $BaSO_4$

Meq of $H_2SO_4$=Meq of $BaSO_4$

Meq of $H_2SO_4=\large\frac{0.5218}{\Large\frac{233.4}{2}}$$\times 1000=4.47$

(E of $BaSO_4=\large\frac{233.4}{2})$

Let $x=$Meq of HCl in the sample

(Meq of HCl+Meq of $H_2SO_4$)=Meq of $NaOH$

$\Rightarrow x+4.47=\large\frac{1}{2}$$\times 24.10\Rightarrow x=7.58$

$\Rightarrow \large\frac{g}{36.5}$$\times 1000=7.58$

$\Rightarrow g=0.2767gm$

This is the wt of HCl in 25ml

Wt of HCl in 1L=$\large\frac{1000}{25}$$\times 0.2767$

$\Rightarrow 11.05g/l$

Similarly,mass of $H_2SO_4$ from this meq=(4.47),in 1L of sample=8.76g/l

Hence (b) is the correct answer.

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