A solution containing ${\mathrm{Na}}_2{\mathrm{CO}}_3$ and NaOH requires 300 ml of 0.1 N HCl using phenolphthalein as an indicator. Methyl orange is then added to the above titrated solution when a further 25 ml of 0.2 N HCl is required. The amount of NaOH present in solution is $\left(\mathrm{NaOH}=40,{\mathrm{Na}}_2{\mathrm{CO}}_3=106\right)$

(I)  Phenolphthalein indicate partial neutralization of 

$\begin{array}{l}{\mathrm{Na}}_2{\mathrm{CO}}_3\to {\mathrm{NaHCO}}_3\\ \mathrm{Meq}.\mathrm{of}{\mathrm{Na}}_2{\mathrm{CO}}_3+\mathrm{Meq}.\mathrm{of}\mathrm{NaOH}=\mathrm{Meq}.\mathrm{of}\mathrm{HCl}\\ \frac{\mathrm{W}}{\mathrm{E}}\times 1000+\frac{\mathrm{W}}{\mathrm{E}}\times 1000=\mathrm{NV}\\ \left(\mathrm{Suppose}{\mathrm{Na}}_2{\mathrm{CO}}_3=\mathrm{a}\mathrm{g},\mathrm{NaOH}=\mathrm{b}\mathrm{g}\right)\\ \frac{\mathrm{a}}{53}\times 1000+\frac{\mathrm{b}}{40}\times 1000=300\times 0.1......\left(1\right)\end{array}$

(II) Methyl orange indicates  complete neutralization

    HCl             HCl     

$\begin{array}{l}{\mathrm{N}}_1{\mathrm{V}}_1={\mathrm{N}}_2{\mathrm{V}}_2,25\times 0.2=0.1\times {\mathrm{V}}_2\mathrm{so}{\mathrm{V}}_2=50\mathrm{ml}\mathrm{excess}\\ \therefore \frac{\mathrm{a}}{53}\times 1000+\frac{\mathrm{b}}{40}\times 1000=350\times 0.1.....\left(2\right)\\ \mathrm{From}\left(1\right)\mathrm{and}\left(2\right)\mathrm{b}=1\mathrm{gm}.\end{array}$