$35.4\mathrm{mL}$ of $\mathrm{HCl}$ is required for the neutralization of a solution containing $0.275 g$ of soluium hydroxide. The normality of hydrochloric acid is

We know that $1g$ equivalent weight of 

                   $\mathrm{NaOH}=40\mathrm{g}$

 $\therefore 40\mathrm{g}$ of $\mathrm{NaOH}=1\mathrm{g}$ eq. of $\mathrm{NaOH}$

      $\begin{array}{r}\therefore 0.275\mathrm{g}\text{ of }\mathrm{NaOH}=\frac{1}{40}\times 0.275\mathrm{eq}.\\ =\frac{1}{40}\times 0.275\times 1000=6.88\mathrm{meq}.\end{array}$

      $\therefore \underset{\left(HCl\right)}{N_1{V}_1}=\underset{\left(NaOH\right)}{N_2{V}_2}$

   $\begin{array}{r}{N}_1\times 35.4=6.88(\because \text{ meq }=NV)\\ N_1=0.194.\end{array}$