Question

A complex is represented as $CoC{l_3}.xN{H_3}$ its 0.1 molal solutions in aq. solution shows $\Delta {T_f} = 0.558^\circ C$ . ${K_f}$ for ${H_2}O$ is $1.86Kmo{l^{ - 1}}kg$ . Assuming $100\%$ ionization of complex and coordination no. of Co is six, calculate the formula of the complex.

Answer 1

For solving the given question, follow these steps: at first calculate the Van't Hoff factor and the depression in freezing point. Then put the value in the equation to find out the number of ions for dissociation. In the end, from the number of ions, write down the formula of the complex.

Formula used: $\Delta {T_f} = \dfrac{{{K_f} \times {w_B} \times 1000}}{{{w_A} \times {M_B}}}$ $i = 1 + \alpha \left( {n - 1} \right)$

Complete step by step answer:
When a substance starts to freeze, the molecules slow down due to the decreases in temperature and mobility of ions; and the intermolecular forces start to take over. At the freezing point of a solvent, an equilibrium exists between the liquid state and the solid-state of the solvent. This tells that the vapor pressures of both the liquid and the solid phase are equal.
On adding non-volatile solute, the vapor pressure of the solution is observed to be lower than the vapor pressure of the pure solvent. This causes the solid and the solution to reach equilibrium at lower temperatures and that causes freezing point depression.
Depression in freezing point can be calculated by using the below formula:
$\Delta {T_f} = \dfrac{{{K_f} \times {w_B} \times 1000}}{{{w_A} \times {M_B}}}$
Where ${K_f}$ = molal depression constant
${w_B}$ = weight of solute
${w_A}$ = weight of solvent
${M_B}$ = molar mass of the solute
The molality is $m = \dfrac{{{w_B} \times 1000}}{{{w_A} \times {M_B}}}$ .
In aqueous solution, $CoC{l_3}.xN{H_3}$ it dissociates $100\%$ .
Hence, upon dissociation let we will have n ions, to correct this anomaly, we will use Van’t Hoff Factor. The Van't Hoff factor for this case can be calculated as:
$i = 1 + \alpha \left( {n - 1} \right)$ where $\alpha$ is the degree of dissociation and n is the number of ions formed as 10% dissociation takes place so $\alpha$ is 1.
$i = 1 + 1\left( {n - 1} \right){\text{ }}$
Now, the Van't Hoff factor can be calculated as:

$$\Delta {T_f} = i \times {K_f} \times m \\\ 0.558 = i \times 1.86 \times 0.1 \\\ i = \dfrac{{0.558}}{{1.86 \times 0.1}} \\\ i = 3 \\\$$

Hence the number of ions can be calculated as:

$$i = 1 + 1\left( {n - 1} \right) \\\ {\text{3}} = 1 + 1\left( {n - 1} \right){\text{ }} \\\ {\text{2}} = 1\left( {n - 1} \right) \\\ n = 3 \\\$$

Therefore, an aqueous solution $CoC{l_3}.xN{H_3}$ will be dissociated into 3 ions. So, the corresponding formula with coordination number six is, $\left[ {CoCl{{\left( {N{H_3}} \right)}_5}} \right]C{l_2}$ .

Note: We see that the molar mass of glucose is almost double of NaCl. Since molar mass has an inverse relation with freezing point depression, NaCl will have just twice the freezing point depression of glucose as the concentration of both the solutes is the same. Hence, we can say that the freezing point depression of 0.1M sodium chloride is nearly twice than that of 0.1M glucose solution.