25 ml of a solution containing HCl and ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is required 10ml of a 1 N NaOH solution for

neutralization. 20ml of the same acid mixture on treating with an excess of ${\text{AgNO}}_3$ gives

0.1435 g of AgCl, then regarding the normality of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ HCl and the normality of which is correct.

No. of eq of HCl in 20ml = no of eq of AgCl formed

           $\therefore$ No. of eq of HCl in 20 ml $\frac{\text{0.1435}}{\text{143.5}}{\text{\hspace{0.17em}=\hspace{0.17em}10}}^{\text{-3}}\therefore \text{\hspace{0.17em}\hspace{0.17em}250\hspace{0.17em}ml\hspace{0.17em}has\hspace{0.17em}}\frac{{\text{10}}^{\text{-3}}\text{\hspace{0.17em}×\hspace{0.17em}25}}{\text{20}}\text{\hspace{0.17em}eq\hspace{0.17em}=\hspace{0.17em}}\frac{\text{5}}{\text{4}}\text{m.eq}$

            Total no of milli equivalents of acids = 10; $\therefore {\text{\hspace{0.17em}no\hspace{0.17em}of\hspace{0.17em}m\hspace{0.17em}eq\hspace{0.17em}of\hspace{0.17em}=\hspace{0.17em}H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{\hspace{0.17em}=\hspace{0.17em}10\hspace{0.17em}-\hspace{0.17em}}\frac{\text{5}}{\text{4}}\text{\hspace{0.17em}=\hspace{0.17em}8.75}$

      ${\text{No\hspace{0.17em}of\hspace{0.17em}H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{\hspace{0.17em}=\hspace{0.17em}}\frac{\text{8.75}}{\text{25}}\text{\hspace{0.17em}=\hspace{0.17em}0.35;\hspace{0.17em}N\hspace{0.17em}of\hspace{0.17em}HCl\hspace{0.17em}=\hspace{0.17em}}\frac{\text{1.25}}{\text{25}}\text{\hspace{0.17em}=\hspace{0.17em}0.05}$