One molecule of $[P t (N H_{3})_{6}] C l_{4}$ gives five ions in solution and required ( $x$ mole) of $A g N O_{3}$ for complete precipitation of chloride ions.
The value of x is :

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Solution

The correct option is C 4
The reaction is:
$[P t (N H_{3})_{6}] C l_{4} \to [P t (N H_{3})_{6}]^{4 +} + 4 C l^{-}$

It gives 4 moles of $C l^{-}$ ions, for complete precipitation of chloride ions, it requires 4 moles of $A g N O_{3}$ as chloride ion precipitate out in the form of $A g C l$ .
Hence, answer is 4.

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