500 𝑚𝐿 of 0.1 𝑀 𝐾𝐶𝑙, 200 𝑚𝐿 of 0.01 𝑀 $NaN{O}_3$ and 500 𝑚𝐿 of 0.1 𝑀 $AgN{O}_3$ was mixed. The molarity of $K^{+},A{g}^{+},C{l}^{-},N{a}^{+},N{O}_3^{-}$ in the solution would be:

1. $\left[K^{+}\right]=0.04\left[A{g}^{+}\right]=0.04\left[N{a}^{+}\right]=0.002\left[C{l}^{-}\right]=0.04\left[N{O}_{3^{-}}\right]=0.42$
   
2. $\left[K^{+}\right]=0.04\left[N{a}^{+}\right]=0.002\left[N{O}_{3^{-}}\right]=0.0433$
3. $\left[K^{+}\right]=0.04\left[A{g}^{+}\right]=0.05\left[N{a}^{+}\right]=0.0025\left[C{l}^{-}\right]=0.05\left[N{O}_{3^{-}}\right]=0.0525$
   
4. $\left[K^{+}\right]=0.05\left[N{a}^{+}\right]=0.0025\left[N{O}_{3^{-}}\right]=0.0525$

The correct option is B $\left[K^{+}\right]=0.04\left[N{a}^{+}\right]=0.002\left[N{O}_{3^{-}}\right]=0.0433$
Calculating the moles of $KCl,NaN{O}_{3}andAgN{O}_3$

$500mLof0.1MKCl,200mLof0.01MNaN{O}_{3}and500mLof0.1MAgN{O}_3$ was mixed.

Thus, the total volume of the solution $=500+200+500=1200ml$

(i) Moles of 𝐾𝐶𝑙 = Molarity × Volume (in 𝐿)
$n_{kcl}=0.1M\times \frac{500}{1000}=0.05mol$
The dissociation of 𝐾𝐶𝑙 is represented as follows:
$KCl\rightleftharpoons K^{+}+C{l}^{-}$
From the reaction, 1 𝑚𝑜𝑙 of 𝐾𝐶𝑙 dissociates into 1 𝑚𝑜𝑙 of $K^{+}$ and 1 𝑚𝑜𝑙 of $C{l}^{-}$.

So, 0.05 𝑚𝑜𝑙 of 𝐾𝐶𝑙 dissociates into 0.05 𝑚𝑜𝑙 of $K^{+}$ and 0.05 𝑚𝑜𝑙 of $C{l}^{-}$.
(ii) Moles of $NaN{O}_3$ = Molarity × Volume (in 𝐿)
$=0.01M\times \frac{200}{1000}=0.002mol$
The dissociation of $NaN{O}_3$ is represented as follows:
$NaN{O}_3\rightleftharpoons N{a}^{+}+N{O}_3^{-}$

From the reaction, 1 𝑚𝑜𝑙 of $NaN{O}_3$ dissociates into 1 𝑚𝑜𝑙 of $N{a}^{+}$ and 1 𝑚𝑜𝑙 of $N{O}_3^{-}$.
So, 0.002 𝑚𝑜𝑙 of $NaN{O}_3$ dissociates into 0.002 𝑚𝑜𝑙 of $N{a}^{+}$ and 0.002 𝑚𝑜𝑙 of $N{O}_3^{-}$.

(iii) Moles of $AgN{O}_3$ = Molarity × Volume (in 𝐿)
$=0.01M\times \frac{500}{1000}L=0.005mol$
The dissociation of $AgN{O}_3$ is represented as follows:
$AgN{O}_3\rightleftharpoons A{g}^{+}+N{O}_3^{-}$

From the reaction, 1 𝑚𝑜𝑙 of $AgN{O}_3$ dissociates into 1 𝑚𝑜𝑙 of $A{g}^{+}$ and 1 𝑚𝑜𝑙 of $N{O}_3^{-}$ .

So, 0.05 𝑚𝑜𝑙 of $AgN{O}_3$ dissociates into 0.05 𝑚𝑜𝑙 of 𝐴𝑔+ and 0.05 𝑚𝑜𝑙 of $N{O}_3^{-}$ .
Calculating the molarity of all the ions present in the solution
$Molesof{K}^{+}=0.05mol$
$MolesofC{l}^{-}=0.05mol$
$MolesofN{a}^{+}=0.002mol$
$MolesofA{g}^{+}=0.05mol$
$MolesofN{O}_3^{-}=0.052mol$
$Molarity=\frac{Numberofmolesofsolute}{volumeofsolution\left(inL\right)}$

$Molarity\left[K^{+}\right]=\frac{0.05}{1200}\times 1000=0.04M$

$Molarity\left[C{l}^{-}\right]=\frac{0.05}{1200}\times 1000=0.04M$

$Molarity\left[N{a}^{+}\right]=\frac{0.002}{1200}\times 1000=0.0016M\approx 0.002M$

$Molarity\left[A{g}^{+}\right]=\frac{0.05}{1200}\times 1000=0.04M$

$Molarity\left[N{O}_3^{-}\right]=\frac{0.052}{1200}\times 1000=0.043M$

So, option (B) is the correct answer.