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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_2]Cl_2`

C

`[Pt(NH_3)_4Cl]Cl_3`

D

`[Pt(NH_3)_4Cl_4]`

Text Solution

Verified by Experts

The correct Answer is:
B
`Delta T_(f) = i xx K_(f) xx m`
`0.0054 = i xx 1.80 xx 0.001`
i = 3
`i = 1 +(n-1) alpha`
`n =3`
`therefore` The correct formula of the compound is `[Pt(NH_3)_4Cl_2]Cl_2`
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