Answer

Verified

21k+ views

**Hint:**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.

Recently Updated Pages

If a wire of resistance R is stretched to double of class 12 physics JEE_Main

The path difference between two waves for constructive class 11 physics JEE_MAIN

What is the difference between solvation and hydra class 11 chemistry JEE_Main

IfFxdfrac1x2intlimits4xleft 4t22Ft rightdt then F4-class-12-maths-JEE_Main

Sodium chloride is purified by passing hydrogen chloride class 11 chemistry JEE_Main

Consider the following oxyanions PO43P2O62SO42MnO4CrO4S2O52S2O72 class 11 chemistry JEE_Main

Other Pages

Assertion Acidic character of group 16 hydrides increases class 11 chemistry JEE_Main

The reaction of Zinc with dilute and concentrated nitric class 12 chemistry JEE_Main

The phase difference between the particles vibrating class 11 physics JEE_Main

What is the pH of 001 M solution of HCl a 1 b 10 c class 11 chemistry JEE_Main

Which of the following brings about dry bleaching A class 11 chemistry JEE_Main

Write short notes on the following carbylamine rea class 12 chemistry JEE_Main