Question
A $$0.001\,molal$$ solution of $$\left[ {Pt{{\left( {N{H_3}} \right)}_4}C{l_4}} \right]$$ 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
A.
$$\left[ {Pt{{\left( {N{H_3}} \right)}_4}C{l_3}} \right]Cl$$
B.
$$\left[ {Pt{{\left( {N{H_3}} \right)}_4}C{l_2}} \right]C{l_2}$$
C.
$$\left[ {Pt{{\left( {N{H_3}} \right)}_4}Cl} \right]C{l_3}$$
D.
$$\left[ {Pt{{\left( {N{H_3}} \right)}_4}C{l_4}} \right]$$
Answer :
$$\left[ {Pt{{\left( {N{H_3}} \right)}_4}C{l_2}} \right]C{l_2}$$
Solution :
$$\eqalign{
& \Delta {T_f} = i.{K_f}.m\,;\,0.0054 = i \times 1.8 \times 0.001 \cr
& i = 3\,\,{\text{so}}\,{\text{it}}\,{\text{is}}\,\,\left[ {Pt{{\left( {N{H_3}} \right)}_4}C{l_2}} \right]C{l_2}. \cr} $$