Question

A $1.2\%$ solution (w/v) of NaCl is isotonic with $7.2\%$ solution (w/v) of glucose. Calculate the degree of ionization and Van't Hoff factor of NaCl.

Answer 1

To answer this question you should recall the concept of osmotic pressure. Osmotic pressure is a colligative property which depends upon the number of moles of solute present in the solution. We shall equate both the values of osmotic pressures of the solvents to calculate the Van’t Hoff factor and then use it to calculate the degree of ionization .
The formula used:
$\Pi = iCRT$
where $\Pi$= Osmotic pressure, $i$= Van’t Hoff Factor, $C$= concentration, $R$=Universal gas constant and $T$=Temperature

Complete step by step answer:
The molar masses of $NaCl$ and glucose are $58.5\;{\text{g/mol}}$ and $180\;{\text{g/mol}}$ respectively. The expression for the osmotic pressure is as follows:
$\Pi = iCRT$
Since two solutions are isotonic, they have equal osmotic pressure.
${\Pi _{\left( {{\text{NaCl}}} \right)}} = {\Pi _{\left( {{\text{glucose}}} \right)}}$.
This means that:
${i_{\left( {{\text{NaCl}}} \right)}}{C_{\left( {{\text{NaCl}}} \right)}}{\text{ = }}{i_{\left( {{\text{glucose}}} \right)}}{C_{\left( {{\text{glucose}}} \right)}}$
Substituting the appropriate values
$\Rightarrow i\left( {NaCl} \right) \times \dfrac{{1.2}}{{58.5}}{\text{ }} = 1 \times \dfrac{{7.2}}{{180}}$
After solving we get $i\left( {NaCl} \right) = 1.95$.
Let $\alpha$ be the degree of ionization of $NaCl.$
$\therefore \alpha = \dfrac{{i - 1}}{{n\prime - 1}}{\text{ }} = \dfrac{{1.95 - 1}}{{2 - 1}}{\text{ }} = 0.95$.
Here, $n\prime \;$ represents the total number of ions produced by dissociation of one molecule of $NaCl$.

Note:
You should know the difference between Association and Dissociation. The Van’t Hoff factor which is an important aspect of colligative properties is defined as the ratio of the concentration of particles formed when a substance is dissolved to the concentration of the substance by mass. In the case of non-electrolytic substances which do not dissociate in water, the value of $i$ is generally 1. But in the case of ionic substances, due to dissociation, the value of $i$ is equal to the total number of ions present in one formula unit of the substance.