Question

A 0.001 molal solution of $\text{ }\left[ \text{Pt(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{4}}}\text{C}{{\text{l}}_{\text{4}}} \right]\text{ }$ in water had a freezing point depression of$\text{0}\text{.005}{{\text{4}}^{\text{0}}}\text{C }$. If $\text{ }{{\text{K}}_{\text{f}}}\text{ }$ for water is$\text{1}\text{.80 }$, the correct formulation of the above molecule is:
A) $\text{ }\left[ \text{Pt(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{4}}}\text{C}{{\text{l}}_{\text{3}}} \right]\text{Cl }$
B) $\text{ }\left[ \text{Pt(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{4}}}\text{C}{{\text{l}}_{\text{3}}} \right]\text{C}{{\text{l}}_{\text{2}}}\text{ }$
C) $\text{ }\left[ \text{Pt(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{4}}}\text{Cl} \right]\text{C}{{\text{l}}_{\text{2}}}\text{ }$
D) $\text{ }\left[ \text{Pt(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{4}}}\text{C}{{\text{l}}_{\text{4}}} \right]\text{ }$

Answer 1

The depression in freezing point $\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{T}}_{\text{f}}}$ is a colligative property. It depends on the amount of solute. If the solute undergoes the association or dissociation in the solution then the difference in the freezing point is stated as:
$\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{T}}_{\text{f}}}\text{ = i m }{{\text{K}}_{\text{f}}}\text{ }$
Where,
$\text{ }{{\text{K}}_{\text{f}}}\text{ }$ is the ebullioscopic constant
‘$i$‘ is the Van’t Hoff’s factor
$\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{T}}_{\text{f}}}\text{ }$ Are difference in the boiling point of the solution and the pure solvent
‘$m$’ is the molality of the solution

Complete step by step solution:
The freezing point, $\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{T}}_{\text{f}}}\text{ }$of a liquid, is the temperature at which the solution solidifies. When a non-volatile solute is added to a liquid, the escaping tendency of a solvent from liquid to solid diminishes whereas the reverse escaping tendency from solid to liquid phase remains unaffected. The solid solvent begins to dissolve.to prevent this and to restore the equilibrium the temperature of the solution is lowered.
The inorganic salts may associate or dissociate into the solution. Therefore, the depression in freezing point is related to the van’t Hoff factor, molality of the solution, and ebullioscopic constant as follows,
$\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{T}}_{\text{f}}}\text{ = i m }{{\text{K}}_{\text{f}}}\text{ }$
We are given the following data:
The cryoscopic constant of the solution is$\text{ }{{\text{K}}_{\text{f}}}\text{ }=\text{ 1}\text{.80 }$,
The molality of a solution is, $\text{ m = 0}\text{.001 molal }$
Depression in freezing point, $\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{T}}_{\text{f}}}\text{ = 0}\text{.005}{{\text{4}}^{\text{0}}}\text{C }$
Compound given is $\text{ }\left[ \text{Pt(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{4}}}\text{C}{{\text{l}}_{\text{4}}} \right]\text{ }$
We have to find the correct formulation of the molecule.
We will solve this question by considering the van’t Hoff factor ‘i’. The depression in freezing point is the difference $\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{T}}_{\text{f}}}\text{ }$between the freezing point of solution (solvent + solute) and that of the pure solvent.
The van’t Hoff factor can be introduced in calculating the observed value of the difference in the boiling point. Let’s first calculate the value of van’t Hoff factor ‘i’ as follows,
$\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{T}}_{\text{f}}}\text{ = i m }{{\text{K}}_{\text{f}}}\text{ }$
Substitute values, we have
$\begin{aligned} & \text{ 0}\text{.0054 = i }\times 0.001\times \text{ 1}\text{.80 } \\\ & \therefore \text{ i = 3 } \\\ \end{aligned}$
Here, the van’t Hoff value is equal to 3.
It suggests that the 1 molecule $\text{ }\left[ \text{Pt(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{4}}}\text{C}{{\text{l}}_{\text{4}}} \right]\text{ }$dissociates into 3 ions in the solution. The coordination complex $\text{ }\left[ \text{Pt(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{4}}}\text{C}{{\text{l}}_{\text{4}}} \right]\text{ }$dissociate into platinate complex and chloride in the solution. The dissociation of $\text{ }\left[ \text{Pt(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{4}}}\text{C}{{\text{l}}_{\text{4}}} \right]\text{ }$is given as follows,
$\text{ }\left[ \text{Pt(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{4}}}\text{C}{{\text{l}}_{\text{3}}} \right]\text{C}{{\text{l}}_{\text{2}}}\text{ }\to \text{ }{{\left[ \text{Pt(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{4}}}\text{C}{{\text{l}}_{3}} \right]}^{\text{2+}}}\text{ + 2C}{{\text{l}}^{-}}\text{ }$
On dissociation, the solution contains the one $\text{ }{{\left[ \text{Pt(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{4}}}\text{C}{{\text{l}}_{3}} \right]}^{\text{2+}}}$ ion and 2 chloride ions. Thus van’t Hoff factor is 3.
Therefore, if a molecule generates three ions in the solution then the molecule must have the formula as $\text{ }\left[ \text{Pt(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{4}}}\text{C}{{\text{l}}_{\text{3}}} \right]\text{C}{{\text{l}}_{\text{2}}}\text{ }$

Hence, (B) is the correct option.

Note: Note that, When solute associates in solution, the van’t Hoff factor is less than 1($\text{ i }< 1$ ).
When solute dissociates in the solution, the Van’t Hoff factor is greater than 1 ($\text{ i }> 1$)
When solute does not dissociate or associate, the van’t Hoff’s factor is equal to 1. ( $\text{ i =} 1$)