The weight of NaCl present  in 250 mL of 0.1 N solution is      

The formula of normality is

$\text{N}=\frac{\text{W}}{\text{GEW}}\times \frac{1000}{\text{V}\left(\text{mL}\right)}$

As per given data,

$\text{N}=0.1\text{ N}$

$\text{V}=250\text{ mL}$

$\text{GEW }=\frac{\text{GMW}}{1}=\frac{58.5}{1}=58.5$

From normality,

$0.1=\frac{\text{w}}{58.5}\times \frac{{\cancel{1000}}^4}{\cancel{250}}$

$\Rightarrow w=1.46g$