500 ml of 0.l M $KCl$ , 200 ml of 0.01 M ${NaNO}_{3}$ and 500 ml of 0.1 M ${AgNO}_{3}$ was mixed. The molarity of $K^{+}, {Ag}^{+}, {Cl}^{-}, {Na}^{+}, {NO}_{3}^{-}$ in the solution would be :

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$\begin{aligned} [K^{+}] = 0.05; [{Na}^{+}] = 0.0025 \\ [{Cl}^{-}] = 0.05; [{NO}_{3}^{-}] = 0.0525 \end{aligned}$

$[K^{+}] = 0.04; [{Na}^{+}] = 0.00166; [{NO}_{3}^{-}] = 0.0433$

$\begin{array}{l} [K^{+}] = 0.04; [{Ag}^{+}] = 0.05; [{Na}^{+}] = 0.0025; \\ [{Cl}^{-}] = 0.05; [{NO}_{3}^{-}] = 0.0525 \end{array}$

$\begin{array}{l} [K^{+}] = 0.04; [{Ag}^{+}] = 0.04; [{Na}^{+}] = 0.002 \\ [{Cl}^{-}] = 0.04; [{NO}_{3}^{-}] = 0.042 \end{array}$

answer is B.

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Detailed Solution

Molarity of the ions can be calculated as,

$\begin{aligned} {\underset{50}{Ag}}^{+} + {\underset{50}{Cl}}^{-} \to AgCl ↓ \end{aligned}$

$[K^{*}] = \frac{50}{1200}; [{Na}^{+}] = \frac{2}{1200}; [{NO}_{3}^{-}] = \frac{52}{1200}$

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