Question

A complex is represented as $CoC{l_3}.xN{H_3}$ its $0.1$ molal solution in aq. Solution shows $\Delta {T_f} = {0.558^0}C.$ K for ${H_2}O$ is $1.86Kmo{l^{ - 1}}kg$. Assuming $100\%$ ionisation of the complex, calculate the number of $N{H_3}$ associated with Co.

Answer 1

The property of decrease in freezing point when some non-volatile solute is dissolved is called depression of freezing point. The depression in freezing point is represented as $\Delta {T_f}$. The extent to which a solute is dissociated or associated can be expressed by the Van't Hoff factor. This is observed when the solution is non ideal, meaning when the solute undergoes association and dissociation.

Complete answer: We know that
$\Delta {T_f} = i{K_f}m$
So, $i = \dfrac{{\Delta {T_f}}}{{{K_f} \times m}}$
‘i’ represent van't hoff factor.
$i > 1$ so it is dissociation
Now the formula for dissociation is;
$i = 1 + \alpha (n - 1)$
$i = 3,n = 3$
When there is dissociation,
The $CoC{l_3}.xN{H_3}$ will dissociate
In this way $2C{l^{ - 1}} + [Co(N{H_3})Cl]$
The oxidation of Co is six. Hence to equalise it in the complex, we will need x=5, so that Co has six as oxidation number.

Additional knowledge: when the molecular mass determined by any of the colligative properties comes out to be different than the theoretically expected value, the substance is said to show abnormal molecular mass. This is observed when the solution is non ideal, meaning when the solute undergoes association and dissociation.

Note:
The various relations as derived above or the colligative properties are applicable only to the solutions of non-electrolytes which do not undergo any dissociation or association into solution. In case of the aqueous solution of electrolytes, inorganic bases and salts which dissociates completely or to a small extent in the solution, the number of particles in the solution increases.