Solutions Question 339
Question: A 0.001 molal solution of $ [Pt{{(NH_3)}4}Cl_4] $ in water had a freezing point depression of $ 0.0054{}^\circ C $ . If $ K{f} $ for water is 1.80, the correct formulation for the above molecule is
Options:
A) $ [Pt{{(NH_3)}_4}Cl_3]Cl $
B) $ [Pt{{(NH_3)}_4}Cl_2]Cl_2 $
C) $ [Pt{{(NH_3)}_4}Cl]Cl_3 $
D) $ [Pt{{(NH_3)}_4}Cl_4] $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \Delta T_{f}=i.K_{f}.m;0.0054=i\times 1.8\times 0.001 $
$ i=3 $ so it is $ [Pt{{(NH_3)}_4}Cl_2]Cl_2. $
