Question

0.1 M solution of NaCl is found to be isotonic with 1.1 % solution of urea. Calculate the apparent degree of ionisation of NaCl.

Answer 1

Hint: Two solutions are called isotonic if they have the same osmotic pressure. To calculate the apparent degree of ionisation of NaCl, you need to first calculate the van’t Hoff factor for NaCl. You should know the formula for calculating osmotic pressure.

Complete Solution :
- Given 0.1 M solution of NaCl this means, we are given molarity of NaCl solution = 0.1 M.
NaCl solution is found to be isotonic with 1.1% of urea solution. We know that two solutions are said to be isotonic when they have the same osmotic pressure.
Formula of calculating osmotic pressure, denoted by $\pi $ is:
$\pi = i\dfrac{n}{V}RT = iCRT$
- Where, $i$ is the van’t Hoff factor for the solute, n is the number of moles, V is volume of solution, R is the gas constant, and T is temperature.

- Now, ${\pi _{NaCl}} = i \times 0.1 \times RT$
Given, 1.1% of urea solution. 1.1 g of urea is present in 100 mL of solution. Molar mass of urea, having chemical formula $N{H_2}CON{H_2}$, is 60 g. Thus,
Number of moles of urea = $\dfrac{{1.1}}{{60}}$
Volume of urea solution = 0.1 L
Urea is a nonelectrolyte, therefore van't Hoff factor, $i = 1$ for urea.

So, ${\pi _{Urea}} = i\dfrac{n}{V}RT = \dfrac{{1.1}}{{60 \times 0.1}}RT$
Since, ${\pi _{NaCl}} = {\pi _{Urea}}$
$i \times 0.1 \times RT = \dfrac{{1.1}}{{60 \times 0.1}}RT$
$\therefore i = 1.83$

- Ionization reaction of NaCl is:
\[\begin{align}
  & \text{Reaction}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ NaCl\to N{{a}^{+}}+C{{l}^{-}} \\
 & \begin{matrix}
   \text{Initial}\ \text{moles} & \ \ \ \ \ \ \ \ \ 1\ \ \ \ \ \ \ \ & 0\ \ \ \ & 0 \\
\end{matrix} \\
 & \begin{matrix}
   \text{Moles}\ \text{at time T}\ \ \ \ & 1-\alpha \ \ \ & \alpha \ \ \ & \alpha \\
\end{matrix} \\
\end{align}\]
Where, $\alpha $ is the degree of ionisation.

- Thus, van’t Hoff factor, $i$ for NaCl:
$i = \dfrac{{{\text{Number}}\;{\text{of}}\;{\text{moles}}\;{\text{after}}\;{\text{ionisation}}}}{{{\text{Number of moles before ionisation}}}} = \dfrac{{1 + \alpha }}{1} = 1 + \alpha $
$1.83 = 1 + \alpha $
$ \Rightarrow \alpha = 0.83$
Hence, the degree of ionisation of NaCl is 0.83.

Note: Van’t Hoff factor predicts the extent of dissociation or association of solutes in the solvents. It is defined as:
$i = \dfrac{{{\text{Observed colligative property}}}}{{{\text{Normal colligative property}}}}$

- Osmotic pressure is one of the colligative properties. Also, keep in mind that the van't Hoff factor, $i = 1$ for urea because it is a nonelectrolyte, shows no association or dissociation.