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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))_(4)Cl_(3)]Cl`

B

`[Pt(NH_(3))_(4)Cl_(2)]Cl_(2)`

C

`[Pt(NH_(3))_(4)Cl]Cl_(3)`

D

`[Pt(NH_(3))_(4)Cl_(4)]`

Text Solution

Verified by Experts

The correct Answer is:
B
`DeltaT_(f)=i xx K_(f)xx m`
`0.0054= i xx 1.80xx0.001`
`i=3`
`i=1+(n-1)alpha`
n = 3
`therefore ` The correct formula of the compound is `[Pt (NH_(3))_(4)Cl_(2)]Cl_(2)`
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