100 ml each of 1M ${\displaystyle AgN{O}_3}$ and 1M NaCl are mixed. The nitrate ion concentration in the resulting solution is

 $AgN{O}_3+ NaCl\to NaN{O}_3 + AgCl$

For $AgN{O}_3$ 

 Molarity = 1M = M₁

Volume  = 100ml = V₁

For NaCl 

Molarity = 1 M = M₂

Volume = 100 ml = V₂

$Milimoles of N{O_3}^{-}= M_1{V}_1 =1\times 100= 100$

$$\begin{gathered} Concentration of N{O_3}^{-} = \frac{millimoles of N{O_3}^{-}}{Total Volume} \\ = \frac{M_1{V}_1}{V_1+V_2} \\ = \frac{100}{200} \\ = 0.5 M \end{gathered}$$                              

Hence, the correct option (B). 0.5M