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
8 Years agoGrade 12
2 Answers
lalit
8 Years ago
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)
=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.
Rishi Sharma
5 Years ago
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.