Question

Compound $PdC{l}_4.6H_{2}O$ is a hydrated complex; $1m$ aqueous solution of it has freezing point $269.28K$. Assuming 100% ionization of the complex, calculate the number of ions furnished by the complex in the solution. ($K_{f}forwater=1.86Kkgmo{l}^{-1}$)

Answer 1

Step 1:Given data:

$$\begin{gathered} Molality=1m \\ {T}_f=269.28K \\ {K}_f=1.86KKgmo{l}^{-1} \end{gathered}$$

Assuming 100% ionization of the complex, the degree of ionization is given as $\mathrm{\alpha }=1$

Step 2: Finding the value of $∆T_f$:

The depression in the freezing point can be calculated as:

$$\begin{gathered} ∆T_f=T_o-T_f \\ =(273.15-269.28)K \\ =3.87K \end{gathered}$$

Where, $∆T_f=$depression in the freezing point

$T_f=$the freezing point of the given solution

$T_o=$standard freezing point

Step 3: Finding the value of i:

The relation between the depression in freezing point and van't Hoffs factor$\left(\mathrm{i}\right)$ can be given as:

$∆T_f=i{K}_{f}m$

Form this, the value of van't Hoffs factor$\left(\mathrm{i}\right)$can be calculated as:

$$\begin{gathered} i=\frac{∆T_f}{K_{f}m} \\ =\frac{3.87}{1.86\times 1} \\ =\frac{3.87}{1.86}=2.0 \end{gathered}$$

Step 4: Finding the value of n:

The relation of van't Hoffs factor can also be given as:

$i=1+(n-1)\alpha$

Where,

$\mathrm{n}=$ the number of ions furnished in the solution due to dissociation

$\mathrm{\alpha }=$the degree of ionization= 1(given)

Hence, the number of ions dissociated can be calculated as:

$$\begin{gathered} n=\frac{i}{\alpha } \\ =\frac{2.0}{1}=2 \end{gathered}$$

Therefore, only 2 ions are furnished by the complex in the solution.