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$PtCl_4.6H_2O$ can exist as a hydrated complex. If 1 molal aqueos solution has a depression in freezing point of $3.72^{\large\circ}$,assuming 100% ionization and $K_f$ of $1.86^{\large\circ}mol^{-1}kg$ which of the following is the complex in question?

$\begin{array}{1 1}(a)\;[Pt(H_2O)_2Cl_4].4H_2O\\(b)\;[Pt(H_2O)_4Cl_2].Cl_2.2H_2O\\(c)\;[Pt(H_2O)_3Cl_3].Cl.3H_2O\\(d)\;[Pt(H_2O)_6].Cl_4\end{array}$

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Answer: $[Pt(H_2O)_3Cl_3].Cl.3H_2O$
Depression of Freezing point, $\Delta\; T_f = i\;K_f\; m$, where $i$ is the van't Hoff factor, $K_f$ is the Freezing Point Depression Constant and $m$ is the molality.
Given drop in temperature, $\Delta\;T_f = 3.72^{\circ}$, $K_f$ of $1.86^{\large\circ}mol^{-1}kg$ and a molality of 1, we get:
$\quad$ $3.72 = i \times 1.86 \times 1 \rightarrow i = 2$
Since $i=2$, the component must be: $[Pt(H_2O)_3Cl_3].Cl.3H_2O$, which is the only one amongst the choices, which produces $i = 2$.
answered Feb 14, 2014 by sreemathi.v
edited Jul 15, 2014 by balaji.thirumalai

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