Question

A 0.001 molal solution of a complex $[M{A_8}]$ in water has the freezing point of -0.0054$^ \circ C$. Assuming 100% ionization of the complex salt and ${K_f}$ for ${H_2}O = 1.86K \cdot {m^{ - 1}}$. Write the correct representation for the complex.
(A) $[M{A_8}]$
(B) $[M{A_7}]A$
(C) $[M{A_6}]{A_2}$
(D) $[M{A_5}]{A_3}$

Answer 1

Hint: The formula of the depression in the freezing point of the solvent in the case of solid solute is
\[\Delta {T_f} = {K_f} \cdot m \cdot i{\text{ }}\]
van’t Hoff factor is the ratio of normal molar mass of the solute to its abnormal molar mass.

Complete step by step answer:
We are given the freezing point of a complex and we need to find the structural formula of the complex.
- We know that addition of a non-volatile solid to water leads to depression in the freezing point of water. The depression in the freezing point ($\Delta {T_f}$) can be calculated as below.
$\Delta {T_f}$ = Freezing point of pure water – Freezing point of the solution
$\Delta {T_f}$ = 0 – (-0.0054) = 0.0054$^ \circ C$

- We can relate the depression in freezing point with the concentration of solute in the following manner.
\[\Delta {T_f} = {K_f} \cdot m \cdot i{\text{ }}....{\text{(1)}}\]
Here, ${K_f}$ is a molal depression constant and it is given $1.86K \cdot {m^{ - 1}}$, m is the molality of the solute and i is the van’t Hoff factor. Putting all the available values in equation (1), we get
\[\Rightarrow 0.0054 = (1.86)(0.001)(i)\]
So, we can write that
\[\Rightarrow i = \dfrac{{0.0054}}{{1.86 \times 0.001}} = 2.9032 \approx 3\]
So, we obtained that the van’t Hoff factor for this complex is 3.
- van’t Hoff factor is the ratio of normal molar mass of the solute to its abnormal molar mass.
- Here, abnormal molar mass corresponds to the mass of species generated due to ionization of the complex. As the normal mass is $\dfrac{1}{3}$ of its abnormal molar mass, we can say that the complex should be giving three ions in its aqueous solution.
- Out of the complexes given in the options, $[M{A_6}]{A_2}$ is the complex which would give three ions upon ionization. i.e. one $[M{A_6}]$ ion and two A ions.
So, the correct answer is “Option C”.

Note: Do not forget that all the salts and complexes which can form ions upon dissolution in the solvent can give abnormal molar masses and this is due to their dissociation into ions. However, association (dimerization) may also occur in some cases like acetic acid in benzene.