A) \[\left[ Pt{{\left( {{H}_{2}}O \right)}_{6}} \right]C{{l}_{4}}\]
B) \[\left[ Pt{{\left( {{H}_{2}}O \right)}_{4}}C{{l}_{2}} \right]C{{l}_{2}}2{{H}_{2}}O\]
C) \[\left[ Pt{{\left( {{H}_{2}}O \right)}_{3}}C{{l}_{3}} \right]C{{l}_{3}}{{H}_{2}}O\]
D) \[\left[ Pt{{\left( {{H}_{2}}O \right)}_{4}}C{{l}_{4}} \right]4{{H}_{2}}O\]
Correct Answer: C
Solution :
\[{{(\Delta {{T}_{f}})}_{cal}}\,\,=\,\,{{K}_{f}}\times m\] \[{{(\Delta {{T}_{f}})}_{cal}}\,\,=\,\,1.86\times 1=1.86\] \[i=\frac{{{(\Delta {{T}_{f}})}_{obs}}}{{{(\Delta {{T}_{f}})}_{cal}}}\,=\,\,\frac{3.72}{1.86}\] \[=2=1+\left( n-1 \right)\alpha \] Hence \[\alpha =1,\] \[\therefore \,\,\,\,\,\,\,\,\,n=2\] \[\therefore \] Two species will be produced from single species which is only possible for \[[Pt{{\left( {{H}_{2}}O \right)}_{3}}C{{l}_{3}}]Cl,3{{H}_{2}}O\]
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