Solutions Question 401
Question: A 0.001 molal solution of [ $Pt( NH_3 )_4Cl_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 ]Cl_2 $
C) $ [ Pt{{( NH_3 )}_4}Cl_2 ]Cl_3 $
D) $ [ Pt{{( NH_3 )}_4}Cl_4 ] $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \Delta T_{f}=imK_{f};,0.0054=i\times 1.8\times 0.001 $
$ i=3. $ So it is $ [ Pt{{( NH_3 )}_4}Cl ]Cl_2 $ .