Question

$0.001$molal solution of $\lbrack Pt(NH_{3})_{4}Cl_{4}\rbrack$ in water had a freezing point depression of $0.0054ºC.$ if $K_{f}$ for water is $1.80$ the correct formula of the compound is

• A. $\lbrack Pt(NH_{3})_{4}Cl_{3}\rbrack Cl$
• B. $\lbrack Pt(NH_{3})_{4}Cl_{4}\rbrack$
• C. $\lbrack Pt(NH_{3})_{4}Cl_{2}\rbrack Cl_{2}$
• D. $\lbrack Pt(NH_{3})_{4}Cl\rbrack Cl_{3}$

Answer 1

Answer: (C)

$\lbrack Pt(NH_{3})_{4}Cl_{2}\rbrack Cl_{2}$

Answer 2

$\Delta T_{f} = i \times K_{f} \times m$

$0.0054 = i \times 1.8 \times 0.001 \Rightarrow i = 3$

Since it give 3 particles on dissociation the correct formula of the molecules is $\lbrack Pt(NH_{3})_{4}Cl_{2}\rbrack Cl_{2}$.