Question

The relative lowering of vapour pressure of a solution is $0.4$ for a solution containing $1$ mol $\mathrm{NaCl}$ in $3$ mol ${\mathrm{H}}_2\mathrm{O}$. What is degree of ionization of $\mathrm{NaCl}$?

Answer 1

Van't Hoff's factor

• The Van't Hoff factor i is the ratio of actual colligative features to theoretical colligative properties.
• It is the difference between the actual or theoretical molar mass and the aberrant or experimental molar mass.
• It is the ratio of total ions left over after dissociation or association to those still present prior to those processes.
• The Van't Hoff factor is greater than one for dissociation, less than one for association, and equal to one for the non-electrolyte solute.
• As we all know, a solute or electrolyte can only dissociate or associate up to a certain degree in a solvent, which is known as the degree of dissociation.

Step 1: Finding mole fraction

• $$\begin{gathered} \mathrm{mole}\mathrm{fraction}=\frac{\mathrm{mass}\mathrm{of}\mathrm{solute}}{\mathrm{mass}\mathrm{of}\mathrm{solution}} \\ \mathrm{mole}\mathrm{fraction}=\frac{1}{1+3} \\ \mathrm{mole}\mathrm{fraction}=\frac{1}{4} \end{gathered}$$

Step 2: Finding Van't Hoff's factor

• relative lowering of vapour pressure =$\mathrm{i}\times \mathrm{mole}\mathrm{fraction}\mathrm{of}\mathrm{solute}$
• $0.4=\mathrm{i}\times \frac{1}{4}$
• Therefore $\mathrm{i}=1.6$

Step 3: Finding Degree of ionization $\left(\mathrm{\alpha }\right)$

• Degree of dissociation/ionisation = $\mathrm{i}-\frac{1}{\mathrm{n}-1}$
• $\mathrm{NaCl}\to {\mathrm{Na}}^{+}+{\mathrm{Cl}}^{-}$
  So, $\mathrm{n}=2$
  $$\begin{gathered} \mathrm{\alpha }=1.6-\frac{1}{2-1} \\ \mathrm{\alpha }=1.6-1 \\ \mathrm{\alpha }=0.6 \end{gathered}$$

Therefore degree of ionisation is $0.6$or $60\%$