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The secondary valence of Pt^(4+) is six. Calculate the number of moles of AgCl participated, when excess of AgNO_3 solution is added to 2L of 0.1M PtCl_4. 4NH_3solution. |
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Answer» Solution :Secondary valence of `Pt^(4+)` is six and it will be SATISFIED by four `NH_(3)` and two `CI^(-)` IONS. The remaining two `CL^(-)` ions satisfy only primary valence. `PtCl_(4).4NH_(3)` can be written as `[Pt(NH_(3))_(4)Cl_(2)]Cl_(2)`. One mole of this complex GIVES 2 moles of chloride ions in the aqueous solution. Number of moles of complex present in 2L of 0.1M `PtCl_(4).4NH_(3)=MxxV=0.1xx2=0.2` Number of moles of chloride ions present in 2L of 0.1M `PtCl_(4).4NH_(3)` solution `= 2 xx0.2 = 0.4`. Number of moles of AgCl precipitated from 2L of 0.1 M `PtCl_(4).4NH_(3)` is 0.4. |
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