Question

# A 0.001 molal solution of $$[Pt (NH_{3})_{4} Cl_{4} ]$$ in water has a freezing point depression of $$0.0054^{0}C$$. If $$K_{f}$$ for water is 1.80 K.kg.mol$$^{-1}$$, the correct formulation of the above molecule is:

A
[Pt(NH3)4Cl3]Cl
B
[Pt(NH3)4Cl2]Cl2
C
[Pt(NH3)4Cl2]Cl3
D
[Pt(NH3)4Cl4]

Solution

## The correct option is B $$[Pt (NH_{3})_{4} Cl_2 ]Cl_{2}$$$$\Delta T_f=i \times K_f \times m$$$$\implies 0.0054=i\times 1.80\times 0.001$$.$$\implies i=\dfrac {0.0054}{1.80\times 0.001}=3$$.$$\implies 1+\alpha (n-1)=3$$, where $$\alpha =1$$.$$\implies n-1=2$$$$\implies n=3$$$$\implies$$ Number of ions in the complex should be 3. So, option B is the correct answer.Chemistry

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