Question
$$0.001\,molal$$ solution of $$Pt{\left( {N{H_3}} \right)_4}C{l_4}$$ in water had a freezing point depression of $${0.0054^ \circ }C.$$ If $${K_f}$$ for water is $$1.80,$$ the correct formula of the compound is
A.
$$\left[ {Pt{{\left( {N{H_3}} \right)}_4}C{l_3}} \right]Cl$$
B.
$$\left[ {Pt{{\left( {N{H_3}} \right)}_4}C{l_4}} \right]$$
C.
$$\left[ {Pt{{\left( {N{H_3}} \right)}_4}C{l_2}} \right]C{l_2}$$
D.
$$\left[ {Pt{{\left( {N{H_3}} \right)}_4}Cl} \right]C{l_3}$$
Answer :
$$\left[ {Pt{{\left( {N{H_3}} \right)}_4}C{l_2}} \right]C{l_2}$$
Solution :
$$\eqalign{
& \Delta {T_f} = i \times {K_f} \times m \cr
& 0.0054 = i \times 1.8 \times 0.001 \Rightarrow i = 3 \cr} $$
Since it gives 3 particles on dissociation the correct formula of the molecule is $$\left[ {Pt{{\left( {N{H_3}} \right)}_4}C{l_2}} \right]C{l_2}.$$