On adding excess of AgNO3 solution into 0.01 mole complex compound PtBr4⋅xNH3, 0.03 moles yellow precipitate was obtained, then value of 'x' is :

PtBr4⋅xNH3+AgNO3 solution ⟶AgBr↓0.01 mole Excess 0.03 moles ∴ The value of x = 5 i.e., complex is [ PtBrNH35Br3

On adding excess of AgNO3 solution into 0.01 mole complex compound PtBr4⋅xNH3, 0.03 moles yellow precipitate was obtained, then value of 'x' is :

PtBr4⋅xNH3+AgNO3 solution ⟶AgBr↓0.01 mole Excess 0.03 moles ∴ The value of x = 5 i.e., complex is [ PtBrNH35Br3

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On adding excess of AgNO₃ solution into 0.01 mole complex compound ${PtBr}_{4} \cdot x {NH}_{3}$ , 0.03 moles yellow precipitate was obtained, then value of 'x' is :

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$\begin{array}{ccc} {PtBr}_{4} \cdot x {NH}_{3} + {AgNO}_{3} solution & ⟶ & AgBr ↓ \\ 0 .01 mole & Excess & 0 .03 moles \end{array}$
$∴$ The value of x = 5
$i.e., complex is [PtBr {({NH}_{3})}_{5}] {Br}_{3}$

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On adding excess of AgNO3 solution into 0.01 mole complex compound PtBr4⋅xNH3, 0.03 moles yellow precipitate was obtained, then value of 'x' is :

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On adding excess of AgNO₃ solution into 0.01 mole complex compound ${PtBr}_{4} \cdot x {NH}_{3}$ , 0.03 moles yellow precipitate was obtained, then value of 'x' is :