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

Options

A. `[Pt(NH_(3))_(4)CI_(3)]CI`
B. `[Pt(NH_(3))_(4)CI_(2)]CI_(2)`
C. `[Pt(NH_(3))_(4)CI]CI_(3)`
D. `[Pt(NH_(3))_(4)CI_(4)]`

Answer

B

Explanation

`DeltaT_(f) = ixxK_(f) xx m`
`0.0054 = ixx 1.80 xx 0.001`
`i=(0.0054)/(0.001 xx1.80) = 3`
Thus, 1 mole of the complex gives 3 mole of ions in solution as follows.
`[Pt(NH_(3))_(4)CI_(2)]CI_(2) overset("water")hArr [Pt(NH_(3))_(4)CI_(2)]^(2+) +2CI^(-)`