Question

A 0.001 molal solution of [Pt(NH₃)₄Cl₄] 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(NH₃)₄ Cl₃]Cl

(b) [Pt(NH₃)₄ Cl₂]Cl₂

(c) [Pt(NH₃)₄ Cl]Cl₃

(d) [Pt(NH₃)₄ Cl₄]

Answer 1

Answer:

(b) [Pt(NH₃)₄ Cl₂]Cl₂

Explanation:

Depression in fz = 54×10-4 =i×kf×m

m=1×10-3

kf=18×10-1

On solving

i=3

A/C to vantoff factor i=1 + \alpha(n – 1)

\alpha=1 find value of n,

n=3 so total no of ion is 3

[Pt(NH)4Cl2]Cl2

All ammonia are within the bracket because it is strong field ligand.

pls mark this as brainlist