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A 0.001 molal solution of [Pt(NH(3))(4)C...

A 0.001 molal solution of `[Pt(NH_(3))_(4)CI_(4)]` in water had a freezing point depression of `0.0054^(@)C`. If `K_(f)` for water is `1.80`, the correct formulation for the above molecule is

A

`[Pt(NH_3)_4Cl_3]Cl`

B

`[Pt(NH_3)_4Cl_4]`

C

`[Pt(NH_3)_4Cl_2]Cl_2`

D

`[Pt(NH_3)_4Cl]Cl_3`

Text Solution

Verified by Experts

The correct Answer is:
C
`DeltaT_f=ixxK_fxxm`
0.0054=`ixx1.8xx0.001 rArr `i=3
Since it gives 3 particles on dissociation the correct formula of the molecule is `[Pt(NH_3)_4Cl_2]Cl_2`
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