A `0.01` `m` solution of `Pt(NH_3)_4Cl_4` in water had the freezing point of `-0.056°C`. Assuming `100%` ionization of

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Q 1280191017.     A `0.01` `m` solution of `Pt(NH_3)_4Cl_4` in water had the freezing point of `-0.056°C`. Assuming `100%` ionization of the complex in solution, find the formula of the complex ion; given `K_f` (water) `= 1.86` `K` `kg` `mol e^(-1).`
A

`[Pt(NH_3)_4Cl_2]^(2+)`

B

`[Pt(NH_3)_2Cl_2]^(2+)`

C

`[Pt(NH_3)_4Cl_2]^(4+)`

D

`[Pt(NH_3)_2Cl_2]`

HINT


Solution
(Provided By a Student and Checked/Corrected by EXXAMM.com Team)

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