A 0.001 molal solution of [Pt(NH3)4Cl4] in water had a freezing point depression of 0.0054oC. If Kf for water is 1.80, the correct formulation for the above molecule is
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a
[Pt(NH3)4Cl3]Cl
b
[Pt(NH3)4Cl2]Cl2
c
[Pt(NH3)4Cl2]Cl3
d
[Pt(NH3)4Cl4]
answer is B.
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Detailed Solution
ΔTf=imkf;0.0054=i×1.8×0.001i=3So it is PtNH34ClCl2