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A 0.001 molal solution if [Pt(NH3)4Cl4] in water had a freezing point depression of 0.0054°C ..If Kf for water is 1.80 then what is the correct formula of the above compound??

Ipsita , 7 Years ago
Grade 12
anser 2 Answers
lalit
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.
Last Activity: 7 Years ago
Rishi Sharma
Dear Student,
Please find below the solution to your problem.

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 + (n – 1) = 3
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.

Thanks and Regards
Last Activity: 5 Years ago
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